Laws of Motion

NEET Physics · 100 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11

Block on smooth inclined wedge with inclination theta. Wedge given acceleration a to right. Condition for block to remain stationary:

A a = g/cosec(theta)

B a = g/cosec(theta)

C a = g*cos(theta)

D a = g*tan(theta)

Correct Answer

Option D

Solution

ma*cos(theta) = mg*sin(theta), so a = g*tan(theta)

Q12

A body of mass M hits wall with velocity V and bounces back. Impulse experienced:

A MV

B 1.5 MV

C 2MV

D zero

Correct Answer

Option C

Solution

Impulse = change in momentum = MV - (-MV) = 2MV

Q13

Person 60 kg in lift 940 kg moving upward with acceleration 1 m/s^2 (g=10 m/s^2). Tension in cable:

A 8600 N

B 9680 N

C 11000 N

D 1200 N

Correct Answer

Option C

Solution

T = (M+m)*(g+a) = 1000*(10+1) = 11000 N

Q14

Block mass m in contact with cart C. Coefficient of static friction mu. Acceleration alpha to prevent block from falling:

A alpha > mg/mu

B alpha > g/(mu*m)

C alpha >= g/mu

D alpha < g/mu

Correct Answer

Option C

Solution

N = m*alpha. mu*N >= mg, so alpha >= g/mu

Q15

Force F = 6i - 8j + 10k acquires acceleration 1 m/s^2. Mass of body:

A 10 kg

B 20 kg

C 10*sqrt(2) kg

D 2*sqrt(10) kg

Correct Answer

Option C

Solution

|F| = sqrt(36+64+100) = sqrt(200) = 10*sqrt(2) N. M = F/a = 10*sqrt(2) kg

Q16

Lift mass 2000 kg, tension 28000 N. Acceleration:

A 4 m/s^2 upwards

B 4 m/s^2 downwards

C 14 m/s^2 upwards

D 30 m/s^2 downwards

Correct Answer

Option A

Solution

F - Mg = Ma. 28000 - 20000 = 2000*a. a = 4 m/s^2 upwards.

Q17

Three forces on body (see figure). Minimum additional force for resultant only in y-direction:

A sqrt(3)/4 N

B sqrt(3) N

C 0.5 N

D 1.5 N

Correct Answer

Option C

Solution

x-components: 1*cos60 + 2*sin30 = 1.5N. 4N force x-component = 4*sin30 = 2N. Additional 0.5N in +x makes net x = 0.

Q18

Roller coaster: riders experience weightlessness at top of hill (radius 20 m). Speed of car:

A 16-17 m/s

B 13-14 m/s

C 14-15 m/s

D 15-16 m/s

Correct Answer

Option C

Solution

mg = mv^2/R, v = sqrt(g*R) = sqrt(10*20) = sqrt(200) = 14.1 m/s, between 14 and 15 m/s

Q19

Sand dropped on conveyer belt at M kg/s. Force to keep belt at constant velocity v:

A Mv/2 newton

B zero

C Mv newton

D 2Mv newton

Correct Answer

Option C

Solution

F = v*(dM/dt) = v*M = Mv newton

Q20

Block B pushed with initial velocity V. Friction coefficient mu. Time to come to rest:

A g*mu/V

B g/V

C V/g

D V/(g*mu)

Correct Answer

Option D

Solution

Retardation a = mu*g. t = V/a = V/(mu*g)

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