If the mass of a body is M on the surface of the earth, the mass of the same body on the surface of the moon will be
🥳 Wohoo! Correct answer
m = constant, W = mg
Mass is an intrinsic property
Mass doesn't change with location
Mass remains M regardless of gravity
Confusing mass and weight
😢 Uh oh! Incorrect answer, Try again
Mass vs weight distinction
m = constant, W = mg
Mass is an intrinsic property
Mass doesn't change with location
Mass remains M regardless of gravity
Confusing mass and weight
Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is I. If the same rod is bent into a ring and its moment of inertia about its diameter is I'. Then the ratio I/I' will be
🥳 Wohoo! Correct answer
I = mr²/k
Rod: I = mL²/12 Ring: I' = mR²/2 = m(L/2π)²/2
Substitute R = L/2π
Simplify to get I/I' = 2π²/3
Wrong substitution of radius
😢 Uh oh! Incorrect answer, Try again
Compare MI for different shapes
I = mr²/k
Rod: I = mL²/12 Ring: I' = mR²/2 = m(L/2π)²/2
Substitute R = L/2π
Simplify to get I/I' = 2π²/3
Wrong substitution of radius
The ratio of angular speed of a second-hand to the hour-hand of a watch is
🥳 Wohoo! Correct answer
ω = dθ/dt
Second hand: 360°/60s = 6°/s Hour hand: 360°/12hr = 360°/43200s
Convert to same time unit
Ratio = (6°/s)/(1/120°/s) = 720:1
Not converting to same time units
😢 Uh oh! Incorrect answer, Try again
Consider time periods
ω = dθ/dt
Second hand: 360°/60s = 6°/s Hour hand: 360°/12hr = 360°/43200s
Convert to same time unit
Ratio = (6°/s)/(1/120°/s) = 720:1
Not converting to same time units
The kinetic energy of a body of mass 4 kg and momentum 6 Ns will be
🥳 Wohoo! Correct answer
K.E. = p²/2m
Use K.E. = p²/2m where p is momentum
K.E. = (6)²/(2×4)
K.E. = 36/8 = 4.5 J
Not using correct KE-momentum relation
😢 Uh oh! Incorrect answer, Try again
Connect momentum and KE through mass
K.E. = p²/2m
Use K.E. = p²/2m where p is momentum
K.E. = (6)²/(2×4)
K.E. = 36/8 = 4.5 J
Not using correct KE-momentum relation
The angle between velocity and acceleration of a particle describing uniform circular motion is
🥳 Wohoo! Correct answer
a = v²/r (centripetal acceleration)
In uniform circular motion, velocity is tangential to the circle
Centripetal acceleration always points toward the center of the circle
The angle between tangential velocity and radial acceleration is always 90°
Confusing uniform circular motion with non-uniform circular motion
😢 Uh oh! Incorrect answer, Try again
Remember that uniform means constant speed, not constant velocity
a = v²/r (centripetal acceleration)
In uniform circular motion, velocity is tangential to the circle
Centripetal acceleration always points toward the center of the circle
The angle between tangential velocity and radial acceleration is always 90°
Confusing uniform circular motion with non-uniform circular motion
A stone of mass 0.05 kg is thrown vertically upwards. What is the direction and magnitude of net force on the stone during its upward motion?
🥳 Wohoo! Correct answer
F = mg
Only force acting is gravity: F = mg
F = 0.05 × 9.8 = 0.49 N
Direction is downward due to gravity
Thinking upward motion needs upward force
😢 Uh oh! Incorrect answer, Try again
Only gravity acts during free fall
F = mg
Only force acting is gravity: F = mg
F = 0.05 × 9.8 = 0.49 N
Direction is downward due to gravity
Thinking upward motion needs upward force
A particle is projected with a velocity v so that its horizontal range is twice the greatest height attained. The horizontal range is
🥳 Wohoo! Correct answer
R = v²sin2θ/g, H = v²sin²θ/2g
For range: R = v²sin2θ/g. For max height: H = v²sin²θ/2g
Given R = 2H, solve: v²sin2θ/g = 2(v²sin²θ/2g)
Solving gives R = 4v²/5g
Not relating range and height equations
😢 Uh oh! Incorrect answer, Try again
Connect range and height through angle
R = v²sin2θ/g, H = v²sin²θ/2g
For range: R = v²sin2θ/g. For max height: H = v²sin²θ/2g
Given R = 2H, solve: v²sin2θ/g = 2(v²sin²θ/2g)
Solving gives R = 4v²/5g
Not relating range and height equations
The ratio of the dimensions of Planck constant and that of moment of inertia has the dimensions of
🥳 Wohoo! Correct answer
h = E/ν, I = mr²
Planck constant h = E/ν = [ML²T⁻²]/[T⁻¹] = [ML²T⁻¹]
Moment of Inertia I = mr² = [M][L²] = [ML²]
h/I = [ML²T⁻¹]/[ML²] = [T⁻¹] (frequency)
Confusing time and frequency dimensions
😢 Uh oh! Incorrect answer, Try again
Always write dimensions and cancel common terms
h = E/ν, I = mr²
Planck constant h = E/ν = [ML²T⁻²]/[T⁻¹] = [ML²T⁻¹]
Moment of Inertia I = mr² = [M][L²] = [ML²]
h/I = [ML²T⁻¹]/[ML²] = [T⁻¹] (frequency)
Confusing time and frequency dimensions
A particle executing SHM has a maximum speed of 0.5 ms⁻¹ and a maximum acceleration of 1.0 ms⁻². The angular frequency of oscillation is
🥳 Wohoo! Correct answer
vmax = ωA, amax = ω²A
Use SHM equations: vmax = ωA and amax = ω²A
From vmax = ωA: 0.5 = ωA and from amax = ω²A: 1.0 = ω²A
Divide equations: amax/vmax = ω = 1.0/0.5 = 2 rad/s
Students might mix up ω and frequency f
😢 Uh oh! Incorrect answer, Try again
Angular frequency = amax/vmax
vmax = ωA, amax = ω²A
Use SHM equations: vmax = ωA and amax = ω²A
From vmax = ωA: 0.5 = ωA and from amax = ω²A: 1.0 = ω²A
Divide equations: amax/vmax = ω = 1.0/0.5 = 2 rad/s
Students might mix up ω and frequency f
Three bodies, a ring (R), a solid cylinder (C) and a solid sphere (S) having the same mass and same radius roll down the inclined plane without slipping. They start from rest if vR, vC, and vS are velocities of respective bodies, then
🥳 Wohoo! Correct answer
PE = KE(trans) + KE(rot)
Compare moments of inertia: Ring (I = MR²), Cylinder (I = ½MR²), Sphere (I = ⅖MR²)
Apply conservation of energy: mgh = ½mv² + ½Iω². Since v = Rω, substitute to get v = √(2gh/(1+I/mR²))
Compare velocities using different I values: Higher I means lower velocity, so vS > vC > vR
Students often neglect role of moment of inertia
😢 Uh oh! Incorrect answer, Try again
Consider both translational and rotational kinetic energies
PE = KE(trans) + KE(rot)
Compare moments of inertia: Ring (I = MR²), Cylinder (I = ½MR²), Sphere (I = ⅖MR²)
Apply conservation of energy: mgh = ½mv² + ½Iω². Since v = Rω, substitute to get v = √(2gh/(1+I/mR²))
Compare velocities using different I values: Higher I means lower velocity, so vS > vC > vR
Students often neglect role of moment of inertia
A 12 kg bomb at rest explodes into two pieces of 4 kg and 8 kg piece is 20 Na, the kinetic energy of the 8 kg piece is
🥳 Wohoo! Correct answer
p = mv, KE = p²/2m
Apply conservation of momentum: Initial momentum = Final momentum = 0, so p₁ + p₂ = 0, where p₂ = 20 Ns
Since momenta are equal and opposite, momentum of 4 kg piece = -20 Ns
Calculate KE of 8 kg piece using K = p²/2m = (20)²/(2×8) = 25 J
Students might forget to use p²/2m formula
😢 Uh oh! Incorrect answer, Try again
Use momentum conservation first, then convert to energy
p = mv, KE = p²/2m
Apply conservation of momentum: Initial momentum = Final momentum = 0, so p₁ + p₂ = 0, where p₂ = 20 Ns
Since momenta are equal and opposite, momentum of 4 kg piece = -20 Ns
Calculate KE of 8 kg piece using K = p²/2m = (20)²/(2×8) = 25 J
Students might forget to use p²/2m formula
Three projectiles A, B and C are projected at an angle of 30°, 45°, 60° respectively. If RA, RB, and RC are ranges of A, B and C respectively, then (velocity of projection is the same for A, B & C)
🥳 Wohoo! Correct answer
R = (u²sin2θ)/g, R₄₅° = u²/g
First, recall the range formula for projectile motion: R = (u²sin2θ)/g where u is the initial velocity and θ is the angle of projection
Calculate for each angle: For 30°: RA = (u²sin60°)/g = (u²×0.866)/g; For 45°: RB = (u²sin90°)/g = u²/g; For 60°: RC = (u²sin120°)/g = (u²×0.866)/g
Compare the values: sin60° = sin120° = 0.866, sin90° = 1. Therefore RA = RC < RB
Students often forget that complementary angles give equal ranges
😢 Uh oh! Incorrect answer, Try again
Range is maximum at 45° and equal for complementary angles (30° and 60°)
R = (u²sin2θ)/g, R₄₅° = u²/g
First, recall the range formula for projectile motion: R = (u²sin2θ)/g where u is the initial velocity and θ is the angle of projection
Calculate for each angle: For 30°: RA = (u²sin60°)/g = (u²×0.866)/g; For 45°: RB = (u²sin90°)/g = u²/g; For 60°: RC = (u²sin120°)/g = (u²×0.866)/g
Compare the values: sin60° = sin120° = 0.866, sin90° = 1. Therefore RA = RC < RB
Students often forget that complementary angles give equal ranges
Maximum acceleration of the train in which a 50 kg box lying on its floor will remain stationary (Given: Coefficient of static friction between the box and the floor of the train is 0.3 and g = 10 ms⁻²)
🥳 Wohoo! Correct answer
f = μN, f = ma
Apply two conditions: (1) Normal force N = mg = 50×10 = 500N, (2) Maximum friction force f = μN = 0.3×500 = 150N
Use Newton's law in horizontal direction: Maximum friction force = mass × acceleration, 150 = 50a
Solve for acceleration: a = 150/50 = 3.0 ms⁻²
Students often forget to equate friction force to ma
😢 Uh oh! Incorrect answer, Try again
Maximum acceleration is limited by friction force available
f = μN, f = ma
Apply two conditions: (1) Normal force N = mg = 50×10 = 500N, (2) Maximum friction force f = μN = 0.3×500 = 150N
Use Newton's law in horizontal direction: Maximum friction force = mass × acceleration, 150 = 50a
Solve for acceleration: a = 150/50 = 3.0 ms⁻²
Students often forget to equate friction force to ma
The component of a vector r along x-axis has a maximum value if
🥳 Wohoo! Correct answer
rx = rcosθ
Start with the component equation: Component along x-axis = rcosθ, where θ is angle with +ve x-axis
Analyze cosθ values: For +ve x-axis (0°): cos0° = 1; For +ve y-axis (90°): cos90° = 0; For -ve y-axis (270°): cos270° = 0; For 45°: cos45° = 1/√2
Conclude that cosθ has maximum value of 1 at θ = 0°, which occurs when vector is along +ve x-axis
Students may think 45° gives maximum projection due to its importance in other contexts
😢 Uh oh! Incorrect answer, Try again
Component of a vector is maximum when the angle between vector and axis is minimum (0°)
rx = rcosθ
Start with the component equation: Component along x-axis = rcosθ, where θ is angle with +ve x-axis
Analyze cosθ values: For +ve x-axis (0°): cos0° = 1; For +ve y-axis (90°): cos90° = 0; For -ve y-axis (270°): cos270° = 0; For 45°: cos45° = 1/√2
Conclude that cosθ has maximum value of 1 at θ = 0°, which occurs when vector is along +ve x-axis
Students may think 45° gives maximum projection due to its importance in other contexts
A body falls freely for 10 sec. Its average velocity during this journey (take g = 10 ms⁻²)
🥳 Wohoo! Correct answer
S = ut + (1/2)gt², v = gt, vₐᵥᵧ = S/t
Begin by identifying the known parameters: Time t = 10 seconds, Acceleration due to gravity g = 10 ms⁻², Initial velocity u = 0 ms⁻¹
Apply the equation for displacement in free fall: S = ut + (1/2)gt². Substitute values: S = 0 + (1/2) × 10 × 10² = 500m
Calculate average velocity using vₐᵥᵧ = S/t = 500/10 = 50 ms⁻¹
Students often confuse average velocity with final velocity
😢 Uh oh! Incorrect answer, Try again
Average velocity equals half of final velocity in free fall with zero initial velocity
S = ut + (1/2)gt², v = gt, vₐᵥᵧ = S/t
Begin by identifying the known parameters: Time t = 10 seconds, Acceleration due to gravity g = 10 ms⁻², Initial velocity u = 0 ms⁻¹
Apply the equation for displacement in free fall: S = ut + (1/2)gt². Substitute values: S = 0 + (1/2) × 10 × 10² = 500m
Calculate average velocity using vₐᵥᵧ = S/t = 500/10 = 50 ms⁻¹
Students often confuse average velocity with final velocity
The value of acceleration due to gravity at a depth of 1600 km is equal to [Radius of earth = 6400 km]
🥳 Wohoo! Correct answer
g_d = g(1 - d/R)
Use formula g_d = g(1 - d/R) where d is depth, R is Earth's radius
Substitute values: g_d = 9.8(1 - 1600/6400)
Calculate: g_d = 9.8(1 - 0.25) = 7.35 ms⁻²
Not using correct formula for gravity variation with depth
😢 Uh oh! Incorrect answer, Try again
Gravity decreases linearly with depth inside Earth
g_d = g(1 - d/R)
Use formula g_d = g(1 - d/R) where d is depth, R is Earth's radius
Substitute values: g_d = 9.8(1 - 1600/6400)
Calculate: g_d = 9.8(1 - 0.25) = 7.35 ms⁻²
Not using correct formula for gravity variation with depth
Two balls are thrown simultaneously in the air. The acceleration of the centre of mass of the two balls when in air
🥳 Wohoo! Correct answer
a_com = F_ext/M_total
Consider only external forces acting on system
In air, only gravity acts as external force
COM acceleration equals g regardless of individual motions
Thinking individual motions affect COM acceleration
😢 Uh oh! Incorrect answer, Try again
Center of mass follows parabolic path under gravity
a_com = F_ext/M_total
Consider only external forces acting on system
In air, only gravity acts as external force
COM acceleration equals g regardless of individual motions
Thinking individual motions affect COM acceleration
A car moving with a velocity of 20 ms⁻¹ is stopped in a distance of 40 m. If the same car is travelling at double the velocity, the distance travelled by it for the same retardation is
🥳 Wohoo! Correct answer
v² = u² + 2as, s = -u²/2a
Use equations of motion: v² = u² + 2as where final velocity v = 0
For first case: 0 = (20)² + 2a(40), calculate acceleration a = -5 m/s²
For second case with u = 40 m/s and same a, use s = -u²/2a = -(40)²/2(-5) = 160 m
Forgetting that stopping distance is proportional to square of initial velocity
😢 Uh oh! Incorrect answer, Try again
Double initial velocity means four times kinetic energy, but same retardation
v² = u² + 2as, s = -u²/2a
Use equations of motion: v² = u² + 2as where final velocity v = 0
For first case: 0 = (20)² + 2a(40), calculate acceleration a = -5 m/s²
For second case with u = 40 m/s and same a, use s = -u²/2a = -(40)²/2(-5) = 160 m
Forgetting that stopping distance is proportional to square of initial velocity
A motor pump lifts 6 tones of water from a well of depth 25m to the first floor of height 35 m from the ground floor in 20 minutes. The power of the pump (in kW) is [g = 10 ms⁻²]
🥳 Wohoo! Correct answer
P = mgh/t
Calculate total height: 25 + 35 = 60m. Convert units: 6 tonnes = 6000 kg, 20 min = 1200 s
Calculate work done: mgh = 6000 × 10 × 60
Calculate power: P = work/time = 3600000/1200 = 3000W = 3kW
Forgetting to convert units to SI system
😢 Uh oh! Incorrect answer, Try again
Remember to convert all units to SI before calculation
P = mgh/t
Calculate total height: 25 + 35 = 60m. Convert units: 6 tonnes = 6000 kg, 20 min = 1200 s
Calculate work done: mgh = 6000 × 10 × 60
Calculate power: P = work/time = 3600000/1200 = 3000W = 3kW
Forgetting to convert units to SI system
A body of mass 50 kg, is suspended using a spring balance inside a lift at rest. If the lift starts falling freely, the reading of the spring balance is
🥳 Wohoo! Correct answer
F = m(a-g) for lift problems
Consider forces in free fall: only gravity acts on body
Apply Newton's law: apparent weight = ma - mg, where a = g in free fall
Since a = g, apparent weight = m(g-g) = 0
Not realizing that apparent weight becomes zero in free fall
😢 Uh oh! Incorrect answer, Try again
In free fall, the body and lift fall with same acceleration g
F = m(a-g) for lift problems
Consider forces in free fall: only gravity acts on body
Apply Newton's law: apparent weight = ma - mg, where a = g in free fall
Since a = g, apparent weight = m(g-g) = 0
Not realizing that apparent weight becomes zero in free fall
If A⃗ = 2î + 3ĵ + 8k̂ is perpendicular to B⃗ = 4ĵ - 4î + αk̂, then the value of 'α' is
🥳 Wohoo! Correct answer
A⃗·B⃗ = a₁b₁ + a₂b₂ + a₃b₃ = 0
For perpendicular vectors, dot product = 0: A⃗·B⃗ = 0
Expand: 2(-4) + 3(4) + 8α = 0
Solve: -8 + 12 + 8α = 0, so α = -1/2
Not considering negative values in dot product calculation
😢 Uh oh! Incorrect answer, Try again
Perpendicular vectors have zero dot product
A⃗·B⃗ = a₁b₁ + a₂b₂ + a₃b₃ = 0
For perpendicular vectors, dot product = 0: A⃗·B⃗ = 0
Expand: 2(-4) + 3(4) + 8α = 0
Solve: -8 + 12 + 8α = 0, so α = -1/2
Not considering negative values in dot product calculation
A substance of mass 49.53 g occupies 1.5 cm³ of volume. The density of the substance (in g cm⁻³) with the correct number of significant figures is
🥳 Wohoo! Correct answer
ρ = m/V
Identify the values given: mass = 49.53 g (4 sig figs), volume = 1.5 cm³ (2 sig figs)
Apply density formula: ρ = m/V = 49.53/1.5
The result is 33.02/10 = 3.302 g/cm³. The answer should have 4 sig figs as per multiplication/division rules
Students often forget sig fig rules in division and multiplication
😢 Uh oh! Incorrect answer, Try again
Count significant figures in given values and apply sig fig rules for division
ρ = m/V
Identify the values given: mass = 49.53 g (4 sig figs), volume = 1.5 cm³ (2 sig figs)
Apply density formula: ρ = m/V = 49.53/1.5
The result is 33.02/10 = 3.302 g/cm³. The answer should have 4 sig figs as per multiplication/division rules
Students often forget sig fig rules in division and multiplication
Moment of inertia of a body about two perpendicular axes X and Y in the plane of the lamina are 20 kg m² respectively. Its moment of inertia about an axis perpendicular to the plane of the lamina and passing through the point of intersection of X and Y axes is
🥳 Wohoo! Correct answer
I_z = I_x + I_y
Apply perpendicular axis theorem: I_z = I_x + I_y
Given I_x = I_y = 20 kg m²
Calculate I_z = 20 + 25 = 45 kg m²
Not applying perpendicular axis theorem
😢 Uh oh! Incorrect answer, Try again
Remember perpendicular axis theorem applies to planar bodies
I_z = I_x + I_y
Apply perpendicular axis theorem: I_z = I_x + I_y
Given I_x = I_y = 20 kg m²
Calculate I_z = 20 + 25 = 45 kg m²
Not applying perpendicular axis theorem
A space station is at a height equal to the radius of the Earth. If 'V_E' is the escape velocity on the surface of the Earth, the same on the space station is _____ times V_E
🥳 Wohoo! Correct answer
v = √(2GM/r)
Use escape velocity formula: v = √(2GM/r)
Note r = 2R at space station
Calculate ratio of velocities
Not considering radius change
😢 Uh oh! Incorrect answer, Try again
Consider gravitational potential
v = √(2GM/r)
Use escape velocity formula: v = √(2GM/r)
Note r = 2R at space station
Calculate ratio of velocities
Not considering radius change
An athlete runs along circular track of diameter 80m...
🥳 Wohoo! Correct answer
s=rθ
Calculate distance=(3/4)×πd=60π m
Calculate displacement=2R=40√2 m
Distance > Displacement verified
Arc vs chord
😢 Uh oh! Incorrect answer, Try again
Consider arc length
s=rθ
Calculate distance=(3/4)×πd=60π m
Calculate displacement=2R=40√2 m
Distance > Displacement verified
Arc vs chord
A man weighing 60 kg is in a lift moving down with an acceleration of 1.8 ms⁻². The force exerted by the floor on him is
🥳 Wohoo! Correct answer
N = m(g-a)
Draw FBD showing all forces: normal force N upward, apparent weight downward. Write N + F = mg where F is pseudo force
Calculate pseudo force F = ma = 60 × 1.8 = 108N downward
Solve N = mg - ma = 60(9.8 - 1.8) = 480N
Forgetting pseudo force
😢 Uh oh! Incorrect answer, Try again
Consider both real and pseudo forces in accelerating frame
N = m(g-a)
Draw FBD showing all forces: normal force N upward, apparent weight downward. Write N + F = mg where F is pseudo force
Calculate pseudo force F = ma = 60 × 1.8 = 108N downward
Solve N = mg - ma = 60(9.8 - 1.8) = 480N
Forgetting pseudo force
Two particles of masses m₁ and m₂ have equal kinetic energies. The ratio of their moments is
🥳 Wohoo! Correct answer
K = p²/2m, p = √(2mK)
Write K.E. equations
Express momentum in terms of K.E.
Find ratio of momenta
Not relating K.E. to momentum
😢 Uh oh! Incorrect answer, Try again
K.E. depends on mass and velocity squared
K = p²/2m, p = √(2mK)
Write K.E. equations
Express momentum in terms of K.E.
Find ratio of momenta
Not relating K.E. to momentum
During inelastic collision between two objects, which of the following quantity always remains conserved?
🥳 Wohoo! Correct answer
Σmv = constant
In inelastic collision, kinetic energy is not conserved
Mechanical energy may change due to deformation
Linear momentum must be conserved (no external force)
Confusing energy conservation
😢 Uh oh! Incorrect answer, Try again
Consider fundamental conservation laws
Σmv = constant
In inelastic collision, kinetic energy is not conserved
Mechanical energy may change due to deformation
Linear momentum must be conserved (no external force)
Confusing energy conservation
The resistance R = V/I where V = (100 ± 5)V and I = (10 ± 0.2)A. The percentage error in R is
🥳 Wohoo! Correct answer
ΔR/R = ΔV/V + ΔI/I
Calculate relative error in V
Calculate relative error in I
Add percentage errors
Not adding relative errors
😢 Uh oh! Incorrect answer, Try again
Remember to add relative errors for division
ΔR/R = ΔV/V + ΔI/I
Calculate relative error in V
Calculate relative error in I
Add percentage errors
Not adding relative errors
An object with mass 5kg is acted upon by a force F = (3i + 4j)N. If its initial velocity at t=0 is v = (6i - 12j)ms⁻¹, the time at which it will just have a velocity along y-axis is
🥳 Wohoo! Correct answer
v = u + at
a = F/m = (0.6i + 0.8j) ms⁻²
vx = 6 + 0.6t, vy = -12 + 0.8t
At t = 10s, vx = 0
Not solving component-wise
😢 Uh oh! Incorrect answer, Try again
Use Newton's second law
v = u + at
a = F/m = (0.6i + 0.8j) ms⁻²
vx = 6 + 0.6t, vy = -12 + 0.8t
At t = 10s, vx = 0
Not solving component-wise
The trajectory of a projectile projected from origin is given by the equation y = x - 2x²/5. The initial velocity of the projectile is
🥳 Wohoo! Correct answer
y = (tanθ)x - (g/2u²cos²θ)x²
Compare with standard equation y = (tanθ)x - (g/2u²cos²θ)x²
Here tanθ = 1 and g/2u²cos²θ = 2/5
Solve for u giving 5 ms⁻¹
Not identifying equation form
😢 Uh oh! Incorrect answer, Try again
Match with projectile equation
y = (tanθ)x - (g/2u²cos²θ)x²
Compare with standard equation y = (tanθ)x - (g/2u²cos²θ)x²
Here tanθ = 1 and g/2u²cos²θ = 2/5
Solve for u giving 5 ms⁻¹
Not identifying equation form
If P, Q and R are physical quantities having different dimensions, which of the following combinations can never be a meaningful quantity?
🥳 Wohoo! Correct answer
[P] = [Q] for subtraction
In (P-Q)/R, P and Q must have same dimensions to subtract
Other options can have valid dimensional combinations
Therefore (P-Q)/R is dimensionally inconsistent
Not checking dimension addition rules
😢 Uh oh! Incorrect answer, Try again
Check dimensional homogeneity
[P] = [Q] for subtraction
In (P-Q)/R, P and Q must have same dimensions to subtract
Other options can have valid dimensional combinations
Therefore (P-Q)/R is dimensionally inconsistent
Not checking dimension addition rules
A satellite is orbiting close to the earth and has a kinetic energy K. The minimum extra kinetic energy required by it to just overcome the gravitation pull of the earth is
🥳 Wohoo! Correct answer
K.E.orbital = ½mv₁², K.E.escape = ½mv₂²
For circular orbit, K.E. = GM₁M₂/2r
For escape velocity, total energy must be zero
Additional K.E. needed equals K
Forgetting escape velocity is √2 times orbital velocity
😢 Uh oh! Incorrect answer, Try again
Compare orbital and escape velocities
K.E.orbital = ½mv₁², K.E.escape = ½mv₂²
For circular orbit, K.E. = GM₁M₂/2r
For escape velocity, total energy must be zero
Additional K.E. needed equals K
Forgetting escape velocity is √2 times orbital velocity
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time 't' is proportional to
🥳 Wohoo! Correct answer
P = ma²t
Power = Force × velocity. For constant acceleration: v = at. Force = ma (constant).
Power = ma × at = ma²t
Therefore, power is directly proportional to t
Not using basic definitions
😢 Uh oh! Incorrect answer, Try again
Remember P = Fv
P = ma²t
Power = Force × velocity. For constant acceleration: v = at. Force = ma (constant).
Power = ma × at = ma²t
Therefore, power is directly proportional to t
Not using basic definitions
Two particles which are initially at rest move towards each other under the action of their mutual attraction. If their speeds are v and 2v at any instant, then the speed of center of mass of the system is
🥳 Wohoo! Correct answer
v_cm = (m₁v₁ + m₂v₂)/(m₁ + m₂)
System is isolated - center of mass velocity remains constant
Initially both particles were at rest, so initial center of mass velocity = 0
By conservation of momentum, COM velocity must remain zero
Forgetting COM velocity remains constant in isolated systems
😢 Uh oh! Incorrect answer, Try again
Apply conservation of linear momentum and initial conditions
v_cm = (m₁v₁ + m₂v₂)/(m₁ + m₂)
System is isolated - center of mass velocity remains constant
Initially both particles were at rest, so initial center of mass velocity = 0
By conservation of momentum, COM velocity must remain zero
Forgetting COM velocity remains constant in isolated systems
One end of a string of length 'l' is connected to a particle of mass 'm' and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed 'v', the net force on the particle (directed towards the centre) is: (T is the tension in the string)
🥳 Wohoo! Correct answer
F = mv²/r
In circular motion, centripetal force is required for circular path. This force is provided by tension in string.
Net force must equal mv²/r where r is radius (here l). Tension in string provides this force.
Therefore, net force = T (tension) which acts as centripetal force.
Not identifying tension as centripetal force
😢 Uh oh! Incorrect answer, Try again
Consider force providing circular motion
F = mv²/r
In circular motion, centripetal force is required for circular path. This force is provided by tension in string.
Net force must equal mv²/r where r is radius (here l). Tension in string provides this force.
Therefore, net force = T (tension) which acts as centripetal force.
Not identifying tension as centripetal force
Rain is falling vertically with a speed of 12 ms⁻¹. A woman rides a bicycles with a speed of 12 ms⁻¹ in east to west direction. What is the direction in which she should hold her umbrella?
🥳 Wohoo! Correct answer
tan θ = v_horizontal/v_vertical
Use vector addition of velocities. Relative velocity of rain with respect to woman is vector sum of rain's velocity and negative of woman's velocity.
For equal speeds (12 ms⁻¹), resultant makes 45° with vertical. Direction is towards east due to woman's motion westward.
Umbrella should be held at 45° towards east to protect from relative rain direction.
Not considering relative motion
😢 Uh oh! Incorrect answer, Try again
Think about relative velocity
tan θ = v_horizontal/v_vertical
Use vector addition of velocities. Relative velocity of rain with respect to woman is vector sum of rain's velocity and negative of woman's velocity.
For equal speeds (12 ms⁻¹), resultant makes 45° with vertical. Direction is towards east due to woman's motion westward.
Umbrella should be held at 45° towards east to protect from relative rain direction.
Not considering relative motion
A cylindrical wire has a mass (0.3±0.003 g), radius (0.5±0.005 mm) and length (6±0.06 cm). The maximum percentage error in the measurement of its density is
🥳 Wohoo! Correct answer
Error = [(Δm/m) + 2(Δr/r) + (Δl/l)]×100%
Density = m/V = m/(πr²l). Percentage error = (Δρ/ρ)×100% = [(Δm/m) + 2(Δr/r) + (Δl/l)]×100%
Calculate each term: Δm/m = 0.003/0.3 = 0.01, Δr/r = 0.005/0.5 = 0.01, Δl/l = 0.06/6 = 0.01
Total = [0.01 + 2(0.01) + 0.01]×100% = 0.04×100% = 4%
Forgetting to double radius error
😢 Uh oh! Incorrect answer, Try again
Remember to double radius error
Error = [(Δm/m) + 2(Δr/r) + (Δl/l)]×100%
Density = m/V = m/(πr²l). Percentage error = (Δρ/ρ)×100% = [(Δm/m) + 2(Δr/r) + (Δl/l)]×100%
Calculate each term: Δm/m = 0.003/0.3 = 0.01, Δr/r = 0.005/0.5 = 0.01, Δl/l = 0.06/6 = 0.01
Total = [0.01 + 2(0.01) + 0.01]×100% = 0.04×100% = 4%
Forgetting to double radius error
At a metro station, a girl walks up a stationary escalator in 20 sec. If she remains stationary on the escalator, then the escalator take her up in 30 sec. The time taken by her to walk up on the moving escalator will be
🥳 Wohoo! Correct answer
1/t₃ = 1/t₁ + 1/t₂
Let v₁ be girl's speed and v₂ be escalator's speed. For stationary escalator: h = v₁t₁, where h is height and t₁ = 20s.
For stationary girl: h = v₂t₂, where t₂ = 30s. For moving escalator: h = (v₁ + v₂)t₃
Using h/v₁ = 20 and h/v₂ = 30, solve for t₃: 1/t₃ = 1/20 + 1/30. Therefore t₃ = 12 sec
Not using reciprocal addition
😢 Uh oh! Incorrect answer, Try again
Consider relative velocities
1/t₃ = 1/t₁ + 1/t₂
Let v₁ be girl's speed and v₂ be escalator's speed. For stationary escalator: h = v₁t₁, where h is height and t₁ = 20s.
For stationary girl: h = v₂t₂, where t₂ = 30s. For moving escalator: h = (v₁ + v₂)t₃
Using h/v₁ = 20 and h/v₂ = 30, solve for t₃: 1/t₃ = 1/20 + 1/30. Therefore t₃ = 12 sec
Not using reciprocal addition
A wheel starting from rest gains angular velocity of 10 rad/s after uniformly accelerated for 5 sec. Total angle turned is
🥳 Wohoo! Correct answer
θ = ωₒt + ½αt²
Use θ = ωₒt + ½αt²
ωₒ = 0, ω = 10 = αt
θ = ½(10)(5) = 25 rad
Forgetting factor of ½
😢 Uh oh! Incorrect answer, Try again
Remember angular motion equations
θ = ωₒt + ½αt²
Use θ = ωₒt + ½αt²
ωₒ = 0, ω = 10 = αt
θ = ½(10)(5) = 25 rad
Forgetting factor of ½
The value of acceleration due to gravity at height 10km from surface is x. At what depth inside earth is g same value x?
🥳 Wohoo! Correct answer
g_h = g(1 - h/R), g_d = g(1 - d/2R)
g_h = g(1 - h/R)
g_d = g(1 - d/2R)
Equate both expressions
Height vs depth variation confusion
😢 Uh oh! Incorrect answer, Try again
Remember g varies differently for height vs depth
g_h = g(1 - h/R), g_d = g(1 - d/2R)
g_h = g(1 - h/R)
g_d = g(1 - d/2R)
Equate both expressions
Height vs depth variation confusion
Two bodies of masses 8kg are placed at vertices A and B of equilateral triangle ABC. A third body of mass 2kg is placed at centroid G. If AG=BG=CG=1m, where should fourth body of mass 4kg be placed so resultant force on 2kg body is zero?
🥳 Wohoo! Correct answer
F = GMm/r²
Use Newton's gravitational law for forces. Due to symmetry, forces from A and B are equal and make 60° with CG.
Their resultant along CG must balance force due to 4kg mass at P: 2(8G/1²)cos30° = 4G/r², where r is PG
Solving gives r = 1/√2 meters
Forgetting vector nature
😢 Uh oh! Incorrect answer, Try again
Force vectors add to zero
F = GMm/r²
Use Newton's gravitational law for forces. Due to symmetry, forces from A and B are equal and make 60° with CG.
Their resultant along CG must balance force due to 4kg mass at P: 2(8G/1²)cos30° = 4G/r², where r is PG
Solving gives r = 1/√2 meters
Forgetting vector nature
A coin placed on a rotating turn table just slips if it is placed at a distance of 4cm from the centre. If the angular velocity of the turn table is doubled it will just slip at a distance of
🥳 Wohoo! Correct answer
f = μN = mrω²
Let's analyze forces: For slipping to occur, friction force equals centripetal force needed: μmg = mrω².
When ω doubles to 2ω, new radius r₂ must satisfy: μmg = mr₂(2ω)².
Comparing equations: r₁ω² = r₂(2ω)², so r₂ = r₁/4 = 1cm
Forgetting ω² dependence
😢 Uh oh! Incorrect answer, Try again
Friction provides centripetal force
f = μN = mrω²
Let's analyze forces: For slipping to occur, friction force equals centripetal force needed: μmg = mrω².
When ω doubles to 2ω, new radius r₂ must satisfy: μmg = mr₂(2ω)².
Comparing equations: r₁ω² = r₂(2ω)², so r₂ = r₁/4 = 1cm
Forgetting ω² dependence
A ball hits the floor and rebounds after an inelastic collision. In this case
🥳 Wohoo! Correct answer
ΣP = constant
In any collision, total momentum of system is always conserved. Individual objects' momenta can change.
Mechanical energy is not conserved in inelastic collision - some energy converts to heat/sound.
When ball hits floor, momentum transfers between ball and Earth, but total remains constant. Option C correctly identifies the conserved quantity.
Confusing energy and momentum
😢 Uh oh! Incorrect answer, Try again
System momentum is always conserved
ΣP = constant
In any collision, total momentum of system is always conserved. Individual objects' momenta can change.
Mechanical energy is not conserved in inelastic collision - some energy converts to heat/sound.
When ball hits floor, momentum transfers between ball and Earth, but total remains constant. Option C correctly identifies the conserved quantity.
Confusing energy and momentum
A 1kg ball moving at 12ms⁻¹ collides with a 2kg ball moving in opposite direction at 24ms⁻¹. If coefficient of restitution is 2/3, their velocities after collision are
🥳 Wohoo! Correct answer
e = -(v₂-v₁)/(u₂-u₁)
Use momentum conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. Also use definition of e: e = -(v₂-v₁)/(u₂-u₁) = 2/3
Substitute values: 1(12) + 2(-24) = v₁ + 2v₂ and v₂-v₁ = -2/3(24-12)
Solve simultaneous equations to get v₁ = -28 m/s, v₂ = -4 m/s
Sign convention errors
😢 Uh oh! Incorrect answer, Try again
Use both momentum and restitution equations
e = -(v₂-v₁)/(u₂-u₁)
Use momentum conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. Also use definition of e: e = -(v₂-v₁)/(u₂-u₁) = 2/3
Substitute values: 1(12) + 2(-24) = v₁ + 2v₂ and v₂-v₁ = -2/3(24-12)
Solve simultaneous equations to get v₁ = -28 m/s, v₂ = -4 m/s
Sign convention errors
A particle starts from origin at t=0s with velocity of 10j ms⁻¹ and moves in x-y plane with constant acceleration of (8i+2j)ms⁻². At instant when x-coordinate is 16m, y-coordinate is
🥳 Wohoo! Correct answer
s = ut + ½at²
Use displacement equations separately for x and y components. For x: x = ut + ½at² where u = 0, a = 8
From x equation: 16 = 4t², so t = 2s
Use this time in y equation: y = 10t + t² = 20 + 4 = 24m
Component mixing
😢 Uh oh! Incorrect answer, Try again
Split motion into components
s = ut + ½at²
Use displacement equations separately for x and y components. For x: x = ut + ½at² where u = 0, a = 8
From x equation: 16 = 4t², so t = 2s
Use this time in y equation: y = 10t + t² = 20 + 4 = 24m
Component mixing
For a projectile motion, the angle between velocity and acceleration is minimum and acute at
🥳 Wohoo! Correct answer
θ = tan⁻¹(vy/vx)
Acceleration is constant (g) and downward. Velocity changes direction continuously.
Angle between v and g is minimum when velocity has maximum downward component - at the highest point of trajectory.
This occurs at only one point - when projectile reaches maximum height and begins descending.
Confusing path points
😢 Uh oh! Incorrect answer, Try again
Consider direction of velocity and g
θ = tan⁻¹(vy/vx)
Acceleration is constant (g) and downward. Velocity changes direction continuously.
Angle between v and g is minimum when velocity has maximum downward component - at the highest point of trajectory.
This occurs at only one point - when projectile reaches maximum height and begins descending.
Confusing path points
The trajectory of projectile is
🥳 Wohoo! Correct answer
y = (tanθ)x - (g/2u²cos²θ)x²
Let's analyze forces on projectile: Only gravity acts in absence of air resistance.
With constant acceleration (g), vertical motion is quadratic while horizontal motion is uniform.
This combination gives parabolic path only in absence of air resistance. With air resistance, path is more complex.
Ignoring air resistance effects
😢 Uh oh! Incorrect answer, Try again
Only gravity affects ideal projectile
y = (tanθ)x - (g/2u²cos²θ)x²
Let's analyze forces on projectile: Only gravity acts in absence of air resistance.
With constant acceleration (g), vertical motion is quadratic while horizontal motion is uniform.
This combination gives parabolic path only in absence of air resistance. With air resistance, path is more complex.
Ignoring air resistance effects
Vernier scale 50 divisions = 49 main scale divisions (0.5mm). Least count is:
🥳 Wohoo! Correct answer
LC = MSD/N
Least count = (1 MSD - 1 VSD)/No. of VSD = (0.5mm - x)/50
One VSD = 49/50 × 0.5mm
Therefore, least count = (0.5 - 0.49)/50 = 0.01mm
Unit conversion errors
😢 Uh oh! Incorrect answer, Try again
Consider division value difference
LC = MSD/N
Least count = (1 MSD - 1 VSD)/No. of VSD = (0.5mm - x)/50
One VSD = 49/50 × 0.5mm
Therefore, least count = (0.5 - 0.49)/50 = 0.01mm
Unit conversion errors
Particle displacement x = t³+3 meters. At zero velocity, displacement is:
🥳 Wohoo! Correct answer
v = dx/dt
Velocity v = dx/dt = 3t². Set v = 0 to find when particle stops
3t² = 0 gives t = 0
At t = 0, x = 0³+3 = 0m. This is displacement when velocity is zero!
Forgetting to substitute t
😢 Uh oh! Incorrect answer, Try again
Find t when v=0, then find x
v = dx/dt
Velocity v = dx/dt = 3t². Set v = 0 to find when particle stops
3t² = 0 gives t = 0
At t = 0, x = 0³+3 = 0m. This is displacement when velocity is zero!
Forgetting to substitute t
The maximum range of a gun on horizontal plane is 16 km. If g = 10 ms⁻², then muzzle velocity of a shell is
🥳 Wohoo! Correct answer
R = u²sin2θ/g
For maximum range, projection angle is 45°. Use range formula: R = (u²sin2θ)/g
At 45°, sin2θ = 1. Therefore, R = u²/g
Substitute R = 16000m, g = 10m/s²: 16000 = u²/10. Solving gives u = 400 m/s
Forgetting angle condition
😢 Uh oh! Incorrect answer, Try again
Maximum range occurs at 45°
R = u²sin2θ/g
For maximum range, projection angle is 45°. Use range formula: R = (u²sin2θ)/g
At 45°, sin2θ = 1. Therefore, R = u²/g
Substitute R = 16000m, g = 10m/s²: 16000 = u²/10. Solving gives u = 400 m/s
Forgetting angle condition
Two masses 5kg, 3kg suspended, system accelerating up 2m/s², g=9.8m/s². T₁ is:
🥳 Wohoo! Correct answer
F = m(g+a)
Forces on upper mass: T₁ up, weight+T₂ down. Net force = ma
For 5kg mass: T₁ - 5g - T₂ = 5a. For 3kg mass: T₂ - 3g = 3a
Solve simultaneous equations: T₁ = (m₁+m₂)(g+a) = 94.4N
Forgetting acceleration effect
😢 Uh oh! Incorrect answer, Try again
Consider both masses' acceleration
F = m(g+a)
Forces on upper mass: T₁ up, weight+T₂ down. Net force = ma
For 5kg mass: T₁ - 5g - T₂ = 5a. For 3kg mass: T₂ - 3g = 3a
Solve simultaneous equations: T₁ = (m₁+m₂)(g+a) = 94.4N
Forgetting acceleration effect
Car on circular track radius 10m, speed 10m/s. Suspended bob length 1m, angle with vertical (radians):
🥳 Wohoo! Correct answer
tanθ = v²/rg
Forces on bob: tension and weight. Net force provides centripetal acceleration
tanθ = v²/(rg) where v is car's speed, r is track radius
θ = arctan(10²/(10×10)) = π/4 radians
Mixing up angles
😢 Uh oh! Incorrect answer, Try again
Think about centripetal force
tanθ = v²/rg
Forces on bob: tension and weight. Net force provides centripetal acceleration
tanθ = v²/(rg) where v is car's speed, r is track radius
θ = arctan(10²/(10×10)) = π/4 radians
Mixing up angles
Objects projected at θ₀ and (90°-θ₀) with same speed. Ratio of max heights:
🥳 Wohoo! Correct answer
h = (u²sin²θ)/(2g)
Max height h = (u²sin²θ)/(2g). Compare heights for θ₀ and (90°-θ₀)
Use trigonometric relation sin(90°-θ) = cosθ
Ratio h₁/h₂ = sin²θ₀/sin²(90°-θ₀) = 2tanθ₀:1
Forgetting trig relations
😢 Uh oh! Incorrect answer, Try again
Use complementary angle properties
h = (u²sin²θ)/(2g)
Max height h = (u²sin²θ)/(2g). Compare heights for θ₀ and (90°-θ₀)
Use trigonometric relation sin(90°-θ) = cosθ
Ratio h₁/h₂ = sin²θ₀/sin²(90°-θ₀) = 2tanθ₀:1
Forgetting trig relations
Angular speed increases 1200rpm to 3120rpm in 16s. Angular acceleration is:
🥳 Wohoo! Correct answer
α = Δω/Δt
First, let's convert everything to standard units. For angular speed, we need rad/s. Remember: rpm to rad/s conversion is: ω = (2π×rpm)/60. So 1200rpm = 125.66 rad/s and 3120rpm = 326.73 rad/s
Now we can treat this like a uniform acceleration problem. Just like linear acceleration is Δv/Δt, angular acceleration is Δω/Δt. Here, Δω = (326.73 - 125.66) rad/s and Δt = 16s
Therefore, α = (326.73 - 125.66)/16 = 4 rad/s². Always check units! We wanted rad/s² and that's what we got.
Forgetting unit conversion
😢 Uh oh! Incorrect answer, Try again
Remember to convert rpm to rad/s before calculating acceleration
α = Δω/Δt
First, let's convert everything to standard units. For angular speed, we need rad/s. Remember: rpm to rad/s conversion is: ω = (2π×rpm)/60. So 1200rpm = 125.66 rad/s and 3120rpm = 326.73 rad/s
Now we can treat this like a uniform acceleration problem. Just like linear acceleration is Δv/Δt, angular acceleration is Δω/Δt. Here, Δω = (326.73 - 125.66) rad/s and Δt = 16s
Therefore, α = (326.73 - 125.66)/16 = 4 rad/s². Always check units! We wanted rad/s² and that's what we got.
Forgetting unit conversion
A chain of length 2m, mass 4kg, 60cm hangs from table. Work to pull chain on table (g=10m/s²):
🥳 Wohoo! Correct answer
W = mgl²/2n
Picture this: initially, 60cm (or 0.6m) of the chain is hanging down creating a U-shape. As we pull it up, each small segment of the chain needs different amounts of work because they start at different heights. This isn't like lifting a single mass!
We need to use integration here. The work done is actually related to the center of mass of the hanging part. For a uniform chain, the center of mass of the hanging part is at half the hanging length. As we pull it up, this changes.
Using the formula W = mgl²/2n where l is hanging length and n is total length, we get: W = (4×10×0.6²)/(2×2) = 3.6 Joules. Notice how we used the initial hanging length (0.6m) in our calculation!
Treating chain as point mass
😢 Uh oh! Incorrect answer, Try again
Think about how the work needed varies for different parts of the chain
W = mgl²/2n
Picture this: initially, 60cm (or 0.6m) of the chain is hanging down creating a U-shape. As we pull it up, each small segment of the chain needs different amounts of work because they start at different heights. This isn't like lifting a single mass!
We need to use integration here. The work done is actually related to the center of mass of the hanging part. For a uniform chain, the center of mass of the hanging part is at half the hanging length. As we pull it up, this changes.
Using the formula W = mgl²/2n where l is hanging length and n is total length, we get: W = (4×10×0.6²)/(2×2) = 3.6 Joules. Notice how we used the initial hanging length (0.6m) in our calculation!
Treating chain as point mass
The centre of mass of an extended body on the surface of the earth and its centre of gravity:
🥳 Wohoo! Correct answer
F = G(Mm/r²) where r varies for large bodies
Let's first understand what center of mass is - it's the point where entire mass of body can be considered concentrated. For example, when you balance a ruler on your finger, you're finding its center of mass.
Now, center of gravity is actually the point where net gravitational force acts. Earth's gravitational field isn't uniform - it varies as 1/r². For small objects like a book or ball, this variation is negligible, but for large objects like skyscrapers, it matters!
This is why for large bodies, these points differ slightly. Imagine a tall building - the top is slightly farther from Earth's center than the bottom, so gravity acts differently. But if our object is tiny compared to Earth's radius (6371 km), we can treat the gravitational field as uniform, making these points coincide.
Many think these points always coincide
😢 Uh oh! Incorrect answer, Try again
Think about why your textbook's center of mass and gravity are same, but a tall building's might differ slightly
F = G(Mm/r²) where r varies for large bodies
Let's first understand what center of mass is - it's the point where entire mass of body can be considered concentrated. For example, when you balance a ruler on your finger, you're finding its center of mass.
Now, center of gravity is actually the point where net gravitational force acts. Earth's gravitational field isn't uniform - it varies as 1/r². For small objects like a book or ball, this variation is negligible, but for large objects like skyscrapers, it matters!
This is why for large bodies, these points differ slightly. Imagine a tall building - the top is slightly farther from Earth's center than the bottom, so gravity acts differently. But if our object is tiny compared to Earth's radius (6371 km), we can treat the gravitational field as uniform, making these points coincide.
Many think these points always coincide
The true length of a wire is 3.678 cm. When the length of this wire is measured using instrument A, the length of the wire is 3.5 cm. When the length of the wire is measured using instrument B, it is found to have length 3.38 cm. Then
🥳 Wohoo! Correct answer
Accuracy=closeness to true value
Compare with true value 3.678 cm
A closer to true value (accuracy)
B has less variation (precision)
Confusing accuracy with precision
😢 Uh oh! Incorrect answer, Try again
Consider accuracy vs precision
Accuracy=closeness to true value
Compare with true value 3.678 cm
A closer to true value (accuracy)
B has less variation (precision)
Confusing accuracy with precision
When a planet revolves around the Sun, in general, for the planet
🥳 Wohoo! Correct answer
L=mvr
Apply Kepler's laws
Angular momentum conserved
Area swept per time constant
Energy variation
😢 Uh oh! Incorrect answer, Try again
Consider conservation laws
L=mvr
Apply Kepler's laws
Angular momentum conserved
Area swept per time constant
Energy variation
A particle is in uniform circular motion. Related to one complete revolution of the particle, which among the statements is incorrect?
🥳 Wohoo! Correct answer
v=rω
Analyze one complete revolution
Speed is constant, not zero
Velocity average is zero
Speed vs velocity
😢 Uh oh! Incorrect answer, Try again
Remember speed vs velocity
v=rω
Analyze one complete revolution
Speed is constant, not zero
Velocity average is zero
Speed vs velocity
A body of mass 10 kg is kept on a horizontal surface. The coefficient of kinetic friction between the body and the surface is 0.5. A horizontal force of 60 N is applied on the body. The resulting acceleration of the body is about
🥳 Wohoo! Correct answer
F=ma
Calculate normal force N=mg=10×10=100N
Find friction force f=μN=0.5×100=50N
Net force=60-50=10N, a=10/10=1ms⁻²
Force resolution
😢 Uh oh! Incorrect answer, Try again
Consider all forces
F=ma
Calculate normal force N=mg=10×10=100N
Find friction force f=μN=0.5×100=50N
Net force=60-50=10N, a=10/10=1ms⁻²
Force resolution
A body is moving along a straight line with initial velocity v₀. Its acceleration a is constant. After t seconds, its velocity becomes v. The average velocity of the body over the given time interval is
🥳 Wohoo! Correct answer
v(avg)=Δs/Δt
Use v=v₀+at
Average velocity=displacement/time
v(avg)=(v²-v₀²)/2at
Average vs instantaneous
😢 Uh oh! Incorrect answer, Try again
Consider motion equations
v(avg)=Δs/Δt
Use v=v₀+at
Average velocity=displacement/time
v(avg)=(v²-v₀²)/2at
Average vs instantaneous
Dimensional formula for activity of radioactive substance is
🥳 Wohoo! Correct answer
A=dN/dt
Activity = Rate of decay = dN/dt
Analyze dimensions: N is dimensionless
Therefore dimensions are [T⁻¹]
Dimension analysis
😢 Uh oh! Incorrect answer, Try again
Consider time rate
A=dN/dt
Activity = Rate of decay = dN/dt
Analyze dimensions: N is dimensionless
Therefore dimensions are [T⁻¹]
Dimension analysis
Value of acceleration due to gravity at height equal to half radius of Earth...
🥳 Wohoo! Correct answer
g'=gR²/(R+h)²
Use g'=gR²/(R+h)²
Substitute h=R/2
Calculate g'=(9.8)(4/9)=4.4ms⁻²
Distance relation
😢 Uh oh! Incorrect answer, Try again
Consider inverse square
g'=gR²/(R+h)²
Use g'=gR²/(R+h)²
Substitute h=R/2
Calculate g'=(9.8)(4/9)=4.4ms⁻²
Distance relation
Among given pair of vectors, the resultant of two vectors can never be 3 units. The vectors are
🥳 Wohoo! Correct answer
R²=a²+b²+2abcosθ
Apply triangle inequality: a-b ≤R≤ a+b
For 4,8 units: 4≤R≤12
3 units not possible in range
Misunderstanding Vector Addition Rules. Forgetting that angle between vectors affects resultant magnitude
😢 Uh oh! Incorrect answer, Try again
Consider vector addition
R²=a²+b²+2abcosθ
Apply triangle inequality: a-b ≤R≤ a+b
For 4,8 units: 4≤R≤12
3 units not possible in range
Misunderstanding Vector Addition Rules. Forgetting that angle between vectors affects resultant magnitude
A ball of mass 0.2 kg is thrown vertically down from a height of 10m. It collides with the floor and loses 50% of its energy and then rises back to the same height. The value of its initial velocity is
🥳 Wohoo! Correct answer
mgh+½mv²=constant
Use energy conservation: mgh₁+½mv₁²=mgh₂
Consider 50% energy loss
Calculate initial velocity: v=√(2gh)=14m/s
Energy conservation
😢 Uh oh! Incorrect answer, Try again
Consider energy loss
mgh+½mv²=constant
Use energy conservation: mgh₁+½mv₁²=mgh₂
Consider 50% energy loss
Calculate initial velocity: v=√(2gh)=14m/s
Energy conservation
The moment of inertia of a rigid body about an axis
🥳 Wohoo! Correct answer
I=∑mr²
Analyze MI formula: I=mr²
Position of axis affects r
Therefore MI depends on axis position
Axis dependence
😢 Uh oh! Incorrect answer, Try again
Consider parallel axis theorem
I=∑mr²
Analyze MI formula: I=mr²
Position of axis affects r
Therefore MI depends on axis position
Axis dependence