Laws of Motion MCQ Questions & Answers in Basic Physics | Physics
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51.
Two blocks $$A$$ and $$B$$ of masses $$3m$$ and $$m$$ respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of $$A$$ and $$B$$ immediately after the string is cut, are respectively
A
$$g,\frac{g}{3}$$
B
$$\frac{g}{3},g$$
C
$$g,g$$
D
$$\frac{g}{3},\frac{g}{3}$$
Answer :
$$\frac{g}{3},g$$
Initially system, is in equilibrium with a total weight of $$4\,mg$$ over spring.
$$\therefore kx = 4\,mg$$
When string is cut at the location as shown above. Free body diagram for $$m$$ is
So, force on mass $$m = mg$$
$$ \Rightarrow $$ Acceleration of mass, $$m = g$$
For mass $$3m;$$ free body diagram is
If $$a =$$ acceleration of block of mass $$3m,$$ then
$$\eqalign{
& {F_{{\text{net}}}} = 4mg - 3mg \cr
& \Rightarrow 3m \cdot {a_A} = mg \cr
& {\text{or}}\,{a_A} = \frac{g}{3} \cr} $$
So, accelerations for blocks $$A$$ and $$B$$ are
$${a_A} = \frac{g}{3}\,\,{\text{and}}\,\,{a_B} = g$$
52.
A rocket with a lift-off mass $$3.5 \times {10^4}\,kg$$ is blasted upwards with an initial acceleration of $$10m/{s^2}.$$ Then the initial thrust of the blast is
A
$$3.5 \times {10^5}\,N$$
B
$$7.0 \times {10^5}\,N$$
C
$$14.0 \times {10^5}\,N$$
D
$$1.75 \times {10^5}\,N$$
Answer :
$$7.0 \times {10^5}\,N$$
As shown in the figure $$F - mg = ma$$
$$\eqalign{
& \therefore F = m\left( {g + a} \right) \cr
& = 3.5 \times {10^4}\left( {10 + 10} \right) \cr
& = 7 \times {10^5}N \cr} $$
53.
A uniform sphere of weight $$W$$ and radius $$5\,cm$$ is being held by a string as shown in the figure. The tension in the string will be :
54.
Five persons $$A,B,C,D$$ and $$E$$ are pulling a cart of mass $$100\,kg$$ on a smooth surface and cart is moving with acceleration $$3\,m/{s^2}$$ in east direction. When person $$A$$ stops pulling, it moves with acceleration $$1\,m/{s^2}$$ in the west direction. When person $$B$$ stops pulling, it moves with acceleration $$24\,m/{s^2}$$ in the north direction. The magnitude of acceleration of the cart when only $$A$$ and $$B$$ pull the cart keeping their directions same as the old directions, is
55.
In the figure shown, the pulleys and strings are massless. The acceleration of the block of mass $$4m$$ just after the system is released from rest is
$$\left( {\theta = {{\sin }^{ - 1}}\frac{3}{5}} \right)$$
The $$FBD$$ of blocks is as shown
From Newton’s second law
$$\eqalign{
& 4mg - 2T\cos \theta = 4mA\,......\left( {\text{i}} \right) \cr
& {\text{and}}\,T - mg = ma\,......\left( {{\text{ii}}} \right) \cr} $$
$$\cos \theta = \frac{4}{5}$$ and from constraint we get
$$a = A\cos \theta \,......\left( {{\text{iii}}} \right)$$
Solving eq. (i), (ii) and (iii)
we get acceleration of block of mass $$4m,A = \frac{{5g}}{{11}}\,{\text{downwards}}{\text{.}}$$
56.
A particle of mass $$0.3kg$$ subject to a force $$F = - kx$$ with $$k = 15 N/m.$$ What will be its initial acceleration if it is released from a point $$20cm$$ away from the origin ?
A
$$15m/{s^2}$$
B
$$3m/{s^2}$$
C
$$10m/{s^2}$$
D
$$5m/{s^2}$$
Answer :
$$10m/{s^2}$$
Mass $$\left( m \right) = 0.3\,kg \Rightarrow F = m.a = 15x$$
$$a = - \frac{{15}}{{0.3}}x = \frac{{150}}{3}x = 50x\,a = 50 \times 0.2 = 10\,m/{s^2}$$
57.
A string of negligible mass going over a clamped pulley of mass $$m$$ supports a block of mass $$M$$ as shown in the figure. The force on the pulley by the clamp is given by
A
$$\sqrt 2 Mg$$
B
$$\sqrt 2 \,mg$$
C
$$\sqrt {{{\left( {M + m} \right)}^2} + {m^2}} g$$
D
$$\sqrt {{{\left( {M + m} \right)}^2} + {M^2}} g$$
At equilibrium $$T = Mg$$
$$F.B.D.$$ of pulley $$ = {F_1} = \left( {m + M} \right)g$$
The resultant force on pulley is
$$\eqalign{
& F = \sqrt {F_1^2 + {T_2}} \cr
& F = \sqrt {{{\left( {m + M} \right)}^2} + {M^2}} g \cr} $$
58.
A balloon with mass $$m$$ is descending down with an acceleration $$a$$ (where, $$a < g$$ ). How much mass should be removed from it so that it starts moving up with an acceleration $$a$$ ?
A
$$\frac{{2ma}}{{g + a}}$$
B
$$\frac{{2ma}}{{g - a}}$$
C
$$\frac{{ma}}{{g + a}}$$
D
$$\frac{{ma}}{{g - a}}$$
Answer :
$$\frac{{2ma}}{{g + a}}$$
When, the balloon is descending down with acceleration $$a$$
So, from free body diagram $$mg - B = m \times a\,......\left( {\text{i}} \right)$$
$$\left[ {B \to {\text{Buoyant force}}} \right]$$
Here, we should assume that while removing same mass the volume of balloon and hence,buoyant force will not change.
Let the new mass of the balloon be $$m'.$$
So, mass removed $$\left( {m - m'} \right)$$
$${\text{So,}}\,B - m'g = m' \times a\,......\left( {{\text{ii}}} \right)$$
Solving Eqs. (i) and (ii),
$$\eqalign{
& mg - B = m \times a \cr
& B - m'g = m' \times a \cr
& mg - m'g = ma + m'a \cr
& \left( {mg - ma} \right) = m'\left( {g + a} \right) \cr
& \Rightarrow m\left( {g - a} \right) = m'\left( {g + a} \right) \cr
& m' = \frac{{m\left( {g - a} \right)}}{{g + a}} \cr} $$
So, mass removed $$ = m - m'$$
$$\eqalign{
& \Rightarrow m\left[ {1 - \frac{{\left( {g - a} \right)}}{{\left( {g + a} \right)}}} \right] = m\left[ {\frac{{\left( {g + a} \right) - \left( {g - a} \right)}}{{\left( {g + a} \right)}}} \right] \cr
& \Rightarrow m\left[ {\frac{{g + a - g + a}}{{g + a}}} \right] \cr
& \Rightarrow \Delta m = \frac{{2ma}}{{g + a}} \cr} $$
59.
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads $$49N,$$ when the lift is stationary. If the lift moves downward with an acceleration of $$5m/{s^2},$$ the reading of the spring balance will be
A
$$24\,N$$
B
$$74\,N$$
C
$$15\,N$$
D
$$49\,N$$
Answer :
$$24\,N$$
For the bag accelerating down
$$\eqalign{
& mg - T = ma \cr
& \therefore T = m\left( {g - a} \right) \cr
& = \frac{{49}}{{10}}\left( {10 - 5} \right) = 24.5\,N \cr} $$
60.
A lift of mass $$1000\,kg$$ is moving upwards with an acceleration of $$1\,m/{s^2}.$$ The tension developed in the string, which is connected to lift is $$\left( {g = 9.8\,m/{s^2}} \right)$$
A
$$9800\,N$$
B
$$10800\,N$$
C
$$11000\,N$$
D
$$10000\,N$$
Answer :
$$10800\,N$$
When, lift move upwards with same acceleration, then according to free body diagram of the left
$$\eqalign{
& T - mg = ma \cr
& {\text{or}}\,T = m\left( {g + a} \right) \cr
& {\text{Given,}}\,m = 1000\,kg,\,a = 1\,m/{s^2}, \cr
& g = 9.8\,m/{s^2} \cr
& {\text{Thus,}}\,T = 1000\left( {9.8 + 1} \right) \cr
& = 1000 \times 10.8 \cr
& = 10800\,N \cr} $$