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
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)
Answer (Detailed Solution Below)
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 ______.
Answer (Detailed Solution Below)
Law of Motion Question 2 Detailed Solution
Concept:
When lift is at rest: T = mg
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
Law of Motion Question 3:
An example of rotational motion is
Answer (Detailed Solution Below)
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
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.
Answer (Detailed Solution Below)
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:
\({R_1} = W + ma = W + \frac{W}{g}a = W\left( {1 + \frac{a}{g}} \right)\)
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?
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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 = mrω2
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 × F1
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?
Answer (Detailed Solution Below)
Law of Motion Question 8 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 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?
Answer (Detailed Solution Below)
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/3Law of Motion Question 10:
Find the magnitude of force exerted by a string on pulley.
Answer (Detailed Solution Below)
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:
The FBD of pulley:
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:
\(F = \sqrt {{{100}^2} + {{100}^2}} = 100\sqrt 2 \;N\)