Work, Power & Energy

$\begin{array}{r l} & u ^ {2} = g L (2 + 3 \sin \phi) = \frac {7}{2} g L \\ & \Rightarrow \quad \sin \phi = \frac {1}{2} \\ & \Rightarrow \quad \theta = 9 0 ^ {\circ} + \phi = 1 2 0 ^ {\circ} \end{array}$
Hence, the correct answer is (C).

The centre of mass of the rope is at a height $\frac{h}{2}$ . So, total work done is the sum of work done to pull the bucket of mass M to a height h and the work done to pull the CM of the rope of mass m to a height $\frac{h}{2}$ . So,
$W = M g h + m g \frac {h}{2}$
$\Rightarrow W = \left(M + \frac {m}{2}\right) g h$
Hence, the correct answer is (B).

Initial velocity = final velocity = 0
But displacement $\Delta x \neq 0$ {∵ $\Delta x = \text{Area under } v-t \text{ graph}$ }
From Work Energy Theorem, $W = \Delta KE = 0$ Hence, the correct answer is (B).

$a = \frac {| \vec {F} |}{m} = \frac {5}{2} = 2. 5 \mathrm{ms} ^ {- 2} \\ \text {Velocity at t = 20 s is v = at} \\ \Rightarrow \quad v = (2. 5) (2 0) = 5 0 \mathrm{ms} ^ {- 1} \\ \Rightarrow \quad \mathrm{K.E.} = \frac {1}{2} (2) (2 5 0 0) \\ \Rightarrow \quad \mathrm{K.E.} = 2 5 0 0 \mathrm{J} \\ \text {Hence, the correct answer is (C).} $

From Work Energy Theorem
$\begin{array}{r l} & {\Delta K E = W _ {\mathrm{net}}} \\ {\Rightarrow} & {K _ {f} - K _ {i} = \int P d t} \\ {\Rightarrow} & {\frac {1}{2} m v ^ {2} = \int_ {0} ^ {2} \biggl (\frac {3}{2} t ^ {2} \biggr) d t} \\ {\Rightarrow} & {v ^ {2} = \biggl (\frac {t ^ {3}}{2} \biggr) \bigg | _ {0} ^ {2}} \\ {\Rightarrow} & {v = 2 \mathrm{ms} ^ {- 1}} \end{array}$
$\left\{\because m = 2 \mathrm{kg} \right\}$
Hence, the correct answer is (B).

Work done by gravitational force is equal to the change in kinetic energy.
Let v be the velocity of particle at point B is v

The horizontal component of velocity remains constant, so
$u \cos \alpha = v \cos \left(\frac {\alpha}{2}\right)$
$\Rightarrow v = \frac {u \cos \alpha}{\cos (\alpha / 2)}$
$\begin{array}{l l} \Rightarrow & W = \frac {1}{2} m v ^ {2} - \frac {1}{2} m (u \cos \alpha) ^ {2} \\ \Rightarrow & W = \frac {1}{2} m \left[ \frac {u ^ {2} \cos^ {2} \alpha}{\cos^ {2} (\alpha / 2)} - u ^ {2} \cos^ {2} \alpha \right] \\ \Rightarrow & W = \frac {1}{2} m u ^ {2} \cos^ {2} \alpha \left[ \sec^ {2} \left(\frac {\alpha}{2}\right) - 1 \right] \\ \text { Since } & 1 + \tan^ {2} \theta = \sec^ {2} \theta \\ \Rightarrow & W = \frac {1}{2} m u ^ {2} \cos^ {2} \alpha \tan^ {2} \left(\frac {\alpha}{2}\right) \\ & \text { Hence, the correct answer is (C). } \end{array}$

Increase in P.E. is the increase in P.E. of C.M.
$\begin{array}{l l} & \Delta U = m g \ell_ {C M} (1 - \cos \theta) \\ \Rightarrow & \Delta U = \bigg (\frac {1 0 0}{1 0 0 0} \bigg) (1 0) \bigg (\frac {1}{2} \bigg) \bigg (1 - \frac {1}{2} \bigg) \\ \Rightarrow & \Delta U = 0. 2 5 \mathrm{J} \\ & \text {Hence, the correct answer is (A).} \end{array}$

$(K. E.) _ {\mathrm{man}} = \frac {1}{2} (\mathrm{K.E.}) _ {\mathrm{boy}} \\ \Rightarrow \quad \frac {1}{2} M v _ {m} ^ {2} = \frac {1}{2} \left[ \frac {1}{2} \left(\frac {M}{2}\right) v _ {b} ^ {2} \right] \\ \Rightarrow \quad v _ {m} = \frac {v _ {b}}{2} \\ \text {Further,} \\ \frac {1}{2} M (v _ {m} + 1) ^ {2} = \frac {1}{2} \left(\frac {M}{2}\right) v _ {b} ^ {2} \\ \Rightarrow \quad v _ {m} + 1 = \frac {v _ {b}}{\sqrt {2}} \\ \text {Solving (1) \& (2), we get} \\ v _ {b} = 2 (\sqrt {2} + 1) = 4.8 2 \mathrm{ms} ^ {- 1} \\ v _ {m} = \sqrt {2} + 1 = 2.4 1 \mathrm{ms} ^ {- 1} $
Hence, the correct answer is (C).

Acceleration $= a = \frac{v_0}{t_0}$
$\begin{array}{r l} & P _ {\mathrm{inst}} = F v _ {t} = (m a) v _ {t} = m \left(\frac {v _ {0}}{t _ {0}}\right) (a t) \\ \Rightarrow & P _ {\mathrm{inst}} = m \left(\frac {v _ {0}}{t _ {0}}\right) ^ {2} t \end{array}$
Hence, the correct answer is (D).

In the initial position, the length of the stretched spring is
$L = \frac {h}{\cos 3 7 ^ {\circ}} = \frac {5 h}{4} = 1. 2 5 h$
So, extension in the spring is
x = L - h = 0.2 5 h
PE stored in the spring is
$U = \frac {1}{2} k x ^ {2} = \frac {1}{2} k (0. 2 5 h) ^ {2} = \frac {k h ^ {2}}{3 2}$
When the spring becomes vertical, it acquires natural length and hence the PE gets converted into kinetic energy of the ring
$\Rightarrow \frac {1}{2} m v ^ {2} = \frac {k h ^ {2}}{3 2}$
$\Rightarrow v = \frac {h}{4} \sqrt {\frac {k}{m}}$
Hence, the correct answer is (D).

When the particle at $x_{1}$ is displaced slightly towards left or right, the force is in the same direction. Hence, the equilibrium is unstable, whereas at $x_{2}$ the force is directed opposite to displacement.
Hence, the correct answer is (B).

Since, $a=\left(\frac{m_{2}-m_{1}}{m_{2}+m_{1}}\right)g$ and $v^{2}-u^{2}=2as$
$\Rightarrow v ^ {2} - 0 ^ {2} = 2 \left(\frac {m _ {2} - m _ {1}}{m _ {2} + m _ {1}}\right) g h$
$\Rightarrow v = \sqrt {2 \left(\frac {m _ {2} - m _ {1}}{m _ {2} + m _ {1}}\right) g h}$
Hence, the correct answer is (D).

Acceleration down the plane is $a = g(\sin \theta - \mu \cos \theta)$ velocity of block at time $t$ is $v = at = gt(\sin \theta - \mu \cos \theta)$ Force of friction is $f = \mu mg \cos \theta$
So, rate of work done by force of friction will be
$P = f v = \mu m g ^ {2} t \cos \theta (\sin \theta - \mu \cos \theta)$
Hence, the correct answer is (C).

Let velocity of block at C be v
By Law of Conservation of Energy, we get
$\frac {1}{2} m v _ {0} ^ {2} + m g r (1 - \cos \theta) = \frac {1}{2} m v ^ {2}$
$\Rightarrow v = \sqrt {v _ {0} ^ {2} + 2 g r (1 - \cos \theta)}$
When block leaves the surface, we have
$N = 0$
$\Rightarrow m g \cos \theta = \frac {m v ^ {2}}{r}$
$\Rightarrow r g \cos \theta = v _ {0} ^ {2} + 2 g r (1 - \cos \theta)$
$\Rightarrow 3 r g \cos \theta = \frac {r g}{4} + 2 r g$
$\Rightarrow 3 \cos \theta = \frac {9}{4}$
$\Rightarrow \cos \theta = \frac {3}{4}$
$\Rightarrow \quad \theta = \cos^ {- 1} \left(\frac {3}{4}\right)$
Hence, the correct answer is (B).

x = Elongation in spring due to mass 10 kg
$x = \frac {m g}{k} = \frac {1 0 \times 1 0}{1 0 0} = 1 \mathrm{m}$
$W _ {F} = \frac {1}{2} \times 1 0 0 \times [ (3) ^ {2} - (1) ^ {2} ] - 1 0 \times 1 0 \times 2 = 2 0 0 \mathrm{J}$
Hence, the correct answer is (A).

By Law of Conservation of Energy
$m g h = \frac {1}{2} m v ^ {2}$
$\Rightarrow v ^ {2} = 2 g h = 2 (1 0) (5)$
$\Rightarrow \quad v = 1 0 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (B).

$F \propto x^{-1/3}$
$\mathrm{i.e.,acceleration} a \propto x ^ {- 1 / 3}$
$\Rightarrow v \frac {d v}{d x} = K x ^ {- 1 / 3}$
$\Rightarrow v ^ {2} \propto x ^ {2 / 3}$
$\Rightarrow v \propto x ^ {1 / 3}$
$\mathrm{Now} P = F v$
$\Rightarrow P \propto x ^ {- 1 / 3} \times x ^ {1 / 3}$
$\Rightarrow P \propto x ^ {0}$
So, power is independent of $x$
Hence, the correct answer is (D).

Since forces are conservative in nature, so the line integral of their sum or their difference in a closed path should also be zero. Hence $W_{1}=W_{2}=0$
Hence, the correct answer is (C).

Three forces are acting on the particle with respect to the sphere
i) pseudo force = ma = mg
$\left\{\mathrm{as} a = g \right\}$
ii) weight = mg
iii) normal reaction = N
Of these first two forces are constant and do work. The third force is not constant but it does not perform any work. Applying Work Energy Theorem, in non-inertial frame, we get
$\frac {1}{2} m v _ {r} ^ {2} = \binom{\text { work done by }}{\text { pseudo force}} + \binom{\text { work done by }}{\text { force of gravity }}$
Pseudo force = mg D

Where $v_{r}$ = speed of particle relative to sphere
$\begin{array}{r l} \Rightarrow & \frac {1}{2} m v _ {r} ^ {2} = m g (A B) + m g (A D) \\ \Rightarrow & \frac {1}{2} v _ {r} ^ {2} = g [ R \sin \theta + R (1 - \cos \theta) ] \\ \Rightarrow & \frac {1}{2} v _ {r} ^ {2} = g R (1 + \sin \theta - \cos \theta) \\ \Rightarrow & v _ {r} = \sqrt {2 g R (1 + \sin \theta - \cos \theta)} \end{array}$
Hence, the correct answer is (D).

$\begin{array}{r l} P _ {\text {output}} & = \frac {(1 0 0) (1 0) (9)}{1 2} \\ & \Rightarrow P _ {\text {output}} = 7 5 0 \mathrm{W} \\ & \text {Since,} \eta = 0. 8 \\ & \Rightarrow 0. 8 = \frac {7 5 0}{P _ {\text {input}}} \\ & \Rightarrow P _ {\text {input}} = \frac {7 5 0}{0 . 8} = 9 3 7. 5 \mathrm{W} \\ & \Rightarrow P _ {\text {input}} \simeq 0. 9 4 \mathrm{kW} \end{array}$
Hence, the correct answer is (C).

Let the initial amplitude decreases to $a_{1}$ to the other side i.e., after the first sweep;
$\binom{\text { Decrease in Elastic }}{\text { Potential Energy }} = \binom{\text { Work Done Against }}{\text { Friction }}$
$\Rightarrow \frac {1}{2} k a ^ {2} - \frac {1}{2} k a _ {1} ^ {2} = \mu m g (a + a _ {1})$
$\Rightarrow \frac {1}{2} k (a + a _ {1}) (a - a _ {1}) = \mu m g (a + a _ {1})$
$\Rightarrow a - a _ {1} = \frac {2 \mu m g}{k}\tag{... (1}$
Similarly $a_{1}-a_{2}=\frac{2\mu mg}{k}$
... (2)
$a _ {n - 1} - a _ {n} = \frac {2 \mu m g}{k}\tag{... (3}$
Adding all the above equations, we get
$a - a _ {n} = \frac {2 n \mu m g}{k}\tag{... (4}$
Now, the block stops when
$\begin{array}{r l} & {\mu m g = k a _ {n}} \\ {\Rightarrow} & {a _ {n} = \frac {\mu m g}{k}} \end{array}$
Substituting in the above equation (4), we get
$\begin{array}{r l} & (2 n + 1) \left(\frac {\mu m g}{k}\right) = a \\ \Rightarrow & (2 n + 1) = \frac {k a}{\mu m g} = \frac {(2 0) (0 . 3)}{(0 . 0 4) (1) (1 0)} = 1 5 \\ \Rightarrow & n = 7 \end{array}$
Hence, the correct answer is (C).

$\begin{array}{l} \eta = \frac {P _ {\text {output}}}{P _ {\text {input}}} \times 100 = \frac {100 \times 10 \times 25}{30,000} \times 100 \\ \Rightarrow \eta = 83.3 \% \\ \text {Hence, the correct answer is (C).} \end{array}$

Speed is maximum at equilibrium, so when the block has descended by x, then both springs apply a force 2kx (upwards) on the block as shown in figure.

For equilibrium, we have
$\begin{array}{l} {2 k x = m g} \\ {x = \frac {m g}{2 k}} \end{array}$
Hence, the correct answer is (A).

Work done = change in potential energy
W = m g b + m g (2 b) = 3 m g b
Hence, the correct answer is (C).

At break up, $F = 0$
$\begin{array}{r l} \Rightarrow & - \frac {d U}{d x} = 0 \mathrm{at} x = x _ {0} \\ \Rightarrow & \frac {d}{d x} (a x ^ {- 1 2} - b x ^ {- 6}) = 0 \\ \Rightarrow & a (- 1 2) x _ {0} ^ {- 1 3} - (b) (- 6) x _ {0} ^ {- 7} = 0 \\ \Rightarrow & x _ {0} ^ {6} = \frac {2 a}{b} \\ \Rightarrow & U (x _ {0}) = \frac {a}{x _ {0} ^ {1 2}} - \frac {b}{x _ {0} ^ {6}} \\ \Rightarrow & U (x _ {0}) = \frac {a}{\left(\frac {2 a}{b}\right) ^ {2}} - \frac {b}{\left(\frac {2 a}{b}\right)} \\ \Rightarrow & U (x _ {0}) = \frac {b ^ {2}}{4 a} - \frac {b ^ {2}}{2 a} \\ \Rightarrow & U (x _ {0}) = - \frac {b ^ {2}}{4 a} \end{array}\tag{say}$
This is the energy with which two atoms are associated. So, dissociation energy equals $\frac{b^{2}}{4a}$ .
Hence, the correct answer is (C).