Law of Motion MCQ Quiz in मल्याळम - Objective Question with Answer for Law of Motion - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Mar 12, 2025

നേടുക Law of Motion ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Law of Motion MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Law of Motion MCQ Objective Questions

Top Law of Motion MCQ Objective Questions

Law of Motion Question 1:

A stone is dropped into a river from a stationary balloon 405 m above water. Another stone is thrown vertically down 1.00 s after the first is dropped. The stones strike the water at the same time. Time initial speed of the second stone is (acceleration due to gravity g = 10 m/s2)

  1. 12.625 m/s
  2. 16.625 m/s
  3. 8.625 m/s
  4. 10.625 m/s

Answer (Detailed Solution Below)

Option 4 : 10.625 m/s

Law of Motion Question 1 Detailed Solution

Concept:

We know Equation of Motion is,

\(x = {v_o}t + \frac{1}{2}a{t^2}\)

where a = acceleration, vo = initial velocity, x = position / height, t = time

Calculation:

Given:

For first stone (vo) = 0, x = 405 m, a = g m/s2 = 10 m/s2

\(405 = {0} + \frac{1}{2}× (10)~{t_1^2}\)

t12 = 81

t1 = 9 sec

Now since the second stone is dropped after 1 sec hence the time taken by the second stone to reach the into river

t2 = 9 - 1 = 8 sec

Let the initial velocity of the second stone is v2o now,

\(405 = ({v_{2o}}× 8) + (\frac{1}{2}× (10)×{8^2})\)

(v2o ×  8) = 405 - 320

v2o = 10.625 m/sec

So initial speed of second stone is 10.625 m/sec.

Law of Motion Question 2:

When a body falls freely under gravitational force, it possesses ______.

  1. Maximum weight
  2. Minimum weight
  3. No weight
  4. No effect on its weight

Answer (Detailed Solution Below)

Option 3 : No weight

Law of Motion Question 2 Detailed Solution

Concept:

When lift is at rest: T = mg

RRB JE ME Emech CH4 - Final 10

When lift is accelerating upward: R1 = TU = mg + ma

When lift is accelerating downward: R2 = TD = mg - ma

Calculation:

The apparent weight of the body is moving downward with acceleration (a) is given by:

Apparent weight = m (g - a)

Under the free fall condition (a) = g

Thus, Apparent weight = 0

Law of Motion Question 3:

An example of rotational motion is

  1. Movement of a car on a straight road
  2. Spinning of earth
  3. Movement of drawer of a table
  4. Motion of earth around the sun

Answer (Detailed Solution Below)

Option 2 : Spinning of earth

Law of Motion Question 3 Detailed Solution

Explanation:

Rotation is the process or act of turning.

Let us see some examples

  • Fan moving in the house
  • Rotation of worm driver over worm gear
  • Group of people holding hands in a circle and walking in the same direction

26 June 1

Translation motion

Movement of the drawer of a table

Linear motion (also called a rectilinear motion)

Movement of a car on a straight road

Revolution

The motion of the earth around the sun

Law of Motion Question 4:

A man weighing W Newton entered a lift which moves with an acceleration of a m/s2. Find the force exerted by the man on the floor of lift when lift is moving downward.

  1. \(W\left( {1 + \frac{a}{g}} \right)\)
  2. \(W\left( {1 - \frac{a}{g}} \right)\)
  3. \(W\left( {1 - \frac{g}{a}} \right)\)
  4. \(W\left( {1 + \frac{g}{a}} \right)\)

Answer (Detailed Solution Below)

Option 2 : \(W\left( {1 - \frac{a}{g}} \right)\)

Law of Motion Question 4 Detailed Solution

When the lift is moving downward:

\({R_1} = W - ma = W - \frac{W}{g}a = W\left( {1 - \frac{a}{g}} \right)\)

When the lift is moving upward:

\({R_1} = W + ma = W + \frac{W}{g}a = W\left( {1 + \frac{a}{g}} \right)\)

RRB JE ME Emech CH4 - Final 10

Law of Motion Question 5:

A boat of mass 300 kg moves according to the equation x = 1.2t2 - 0.2t3. When the force will become zero?

  1. 2 s
  2. 1 s
  3. 6 s
  4. 2.8 s

Answer (Detailed Solution Below)

Option 1 : 2 s

Law of Motion Question 5 Detailed Solution

Explanation:

From Newton's second law of motion, the force acting on a body is given by 

F = ma

If force is zero then its acceleration must be zero

F = 0 ⇒ a = 0

\(v=\frac{{{d}}x}{d{{t}}}\)

\(a=\frac{{{d}}v}{d{{t}}}= \frac{{{d}{^2}}x}{d{{t^{2}}}}\)

Now,

If a = 0,then \(\frac{{{d}^{2}}x}{d{{t}^{2}}}=0\)

x = 1.2t2 - 0.2t3

\(\frac{dx}{dt}=2.4t-0.6{{t}^{2}}\)

\(\frac{{{d}^{2}}x}{d{{t}^{2}}}=2.4-1.2~t=0\)

t = 2.4/1.2

∴ t = 2 s

Law of Motion Question 6:

During inelastic collision of two particles which one of the following is conserved?

  1. Total linear momentum only
  2. Total kinetic energy only
  3. Both (1) and (2)
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : Total linear momentum only

Law of Motion Question 6 Detailed Solution

Explanation:

  • Momentum is conserved in all collisions.
  • In elastic collision, kinetic energy is also conserved.
  • In inelastic collision, kinetic energy is not conserved. In perfectly inelastic collision, objects stick together after collision.
  • Hence, In inelastic collision of two particles total linear momentum only is conserved.

Perfectly elastic collision:

If law of conservation of momentum and that of kinetic energy hold good during the collision.

Inelastic collision:

If law of conservation of momentum holds good during collision while that of kinetic energy is not.

Coefficient of restitution (e)

\(e = \frac{{Relative\;velocity\;after\;collision}}{{Relative\;velocity\;before\;collision}} = \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}\)

  • For perfectly elastic collision, e = 1
  • For inelastic collision, e < 1
  • For perfectly inelastic collision, e = 0

Law of Motion Question 7:

If the radius is doubled and angular speed is reduced to half of its original value, the Centrifugal force relative to its original value is

  1. 2 times
  2. 0.5 times
  3. 0.1 times
  4. 0.2 times

Answer (Detailed Solution Below)

Option 2 : 0.5 times

Law of Motion Question 7 Detailed Solution

Concept:

Centrifugal force: The imaginary force acting on the particle moving on a circular path after observing with respect to the particle itself is called centrifugal force.

\(Centrifugal\;force\;\left( F \right) =~\frac{mV_c^2}{r}\)

Also, Vc = rω

∴ F = mrω

where m is mass of the particle, Vc = velocity, r = radius, ω = angular velocity 

Explanation:

Given:

r2 = 2r1\(ω_2~=~\frac{ω_1 }{2}\)

Therefore, by using the above value we have,

\(\frac{F_2}{F_1}~=~\frac{mr_2\omega_2^2}{mr_1\omega_1^2}~=~\frac{2r_1(\frac{\omega_1}{2})^2}{r_1\omega_1^2}\)

F2 = 0.5 × F

Law of Motion Question 8:

A railway engine of mass 60 tonnes is moving in a circular track of radius 200 m with a velocity of 36 kmph. What is the force exerted on the rails towards the centre of the circle?

  1. 30 kN
  2. 180 kN
  3. 3 kN
  4. 300 kN

Answer (Detailed Solution Below)

Option 1 : 30 kN

Law of Motion Question 8 Detailed Solution

Concept:

 

F1 J.K 2.6.20 Pallavi D1

  • Centrifugal force: The imaginary force acting on the particle moving on a circular path after observing with respect to the particle itself is called centrifugal force.
  • Centrifugal force is equal to centripetal force and the direction of centrifugal force is in a radially outward direction

\(Centrifugal\;force\;\left( F \right) = m\frac{{V^2}}{R}\)

where m is mass of the rail.

Calculation:

Given:

m = 60 ton = 60 × 103 kg, V = 36 kmph = 10 m/s, R = 200 m

Therefore, \(F=\frac{60~\times~10^3~\times~10^2}{200}=30~\times~10^3~N=30~kN\)

Law of Motion Question 9:

The tension in the cable supporting a lift moving upwards is twice the tension when the lift moves downwards. What is the acceleration of the lift?

  1. g/4
  2. g/3
  3. g/2
  4. g

Answer (Detailed Solution Below)

Option 2 : g/3

Law of Motion Question 9 Detailed Solution

Concept:

Let lift is moving in an upward direction with acceleration ‘a’.

Tension in the string (T1)  = m(g+a) where m is the mass of lift.

Let lift is moving in a downward direction with acceleration ‘a’.

Tension in the string (T2) = m(g-a)

T1 = 2T2  (given)

m(g+a) = 2m (g-a)

a = g/3

Law of Motion Question 10:

Find the magnitude of force exerted by a string on pulley.

RRB JE ME Emech CH4 - Final 3

  1. 100 N
  2. 200 N
  3. 100√2 N
  4. 2√100 N

Answer (Detailed Solution Below)

Option 3 : 100√2 N

Law of Motion Question 10 Detailed Solution

Concept:

  • The tension in the string will always be away from the system.
  • Since string is massless so, tension in both sides of string will be same.

The FBD of Block of 10 kg:

RRB JE CM 4.1

The FBD of pulley:

RRB JE ME Emech CH4 - Final 5

Calculation:

Given:

m = 10 kg

From the FBD of the Puley, the reultant of both the tension forces on the pulley (which are perpendicular to each other) will be the magnitude of force exerted by a string on pulley.

Force exerted by string:

\(F = \sqrt {{{100}^2} + {{100}^2}} = 100\sqrt 2 \;N\)

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