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

Concept:

We know Equation of Motion is,

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

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

Calculation:

Given:

For first stone (v_o) = 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 v_2o now,

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

(v_2o ×  8) = 405 - 320

v_2o = 10.625 m/sec

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

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

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

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:

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

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

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

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, V_c = rω

∴ F = mrω² 

where m is mass of the particle, V_c = 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 

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

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)

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:

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