1. If the rotational kinetic energy of a body is E and moment of inertia is I, then angular momentum will be: A) EI B) sqrt(2EI) C) E/I D) sqrt(IE) 2. The displacement versus... 1. If the rotational kinetic energy of a body is E and moment of inertia is I, then angular momentum will be: A) EI B) sqrt(2EI) C) E/I D) sqrt(IE) 2. The displacement versus time graph of particles in SHM is shown in figure. The maximum velocity of the particle is: A) 1 cm/s B) 0.2 cm/s C) 3.14 cm/s D) 31.4 cm/s 3. Two capillary tubes P and Q of radii 4 mm and 2 mm are dipped in water. What is the ratio of heights through which liquid rises in the tubes P and Q? A) 1:2 B) 2:1 C) 1:4 D) 4:1 4. A system undergoes an isochoric process so that its temperature changes from 127°C to 227°C. What is the ratio of initial pressure to final pressure? A) 4/5 B) 5/4 C) 227/127 D) 127/227 5. Given figure represents the block diagram of a heat engine. What is the efficiency of this engine? A) 25% B) 50% C) 60% D) 75% Read more

Steps to Solve

1. Question 1: Rotational Kinetic Energy and Angular Momentum

The question relates rotational kinetic energy ($E$), moment of inertia ($I$), and angular momentum ($L$). The formula relating these is $E = \frac{L^2}{2I}$. We need to solve for $L$.

2. Solve for L $E = \frac{L^2}{2I}$ implies $L^2 = 2EI$. Therefore, $L = \sqrt{2EI}$.

3. Question 2: Simple Harmonic Motion (SHM) - Maximum Velocity

The maximum velocity ($v_{max}$) in SHM is given by $v_{max} = A\omega$, where $A$ is the amplitude and $\omega$ is the angular frequency. From the graph, the amplitude $A = 1$ cm. The period $T = 0.2$ s. Therefore, $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi \approx 31.4$ rad/s.

4. Calculate max velocity $v_{max} = A\omega = (1 \text{ cm})(10\pi \text{ rad/s}) = 10\pi \text{ cm/s} \approx 31.4 \text{ cm/s}$.

5. Question 3: Capillary Action

The height ($h$) to which a liquid rises in a capillary tube is inversely proportional to the radius ($r$) of the tube, i.e., $h \propto \frac{1}{r}$. Therefore, $\frac{h_P}{h_Q} = \frac{r_Q}{r_P}$.

6. Calculate Ratio Given $r_P = 4$ mm and $r_Q = 2$ mm, the ratio $\frac{h_P}{h_Q} = \frac{2}{4} = \frac{1}{2}$.

7. Question 4: Isochoric Process

In an ischoric process, volume is constant, and $\frac{P}{T} = \text{constant}$. Therefore, $\frac{P_1}{T_1} = \frac{P_2}{T_2}$, which means $\frac{P_1}{P_2} = \frac{T_1}{T_2}$. We are given $T_1 = 127^\circ C = 127 + 273 = 400$ K and $T_2 = 227^\circ C = 227 + 273 = 500$ K.

8. Calculate Pressure Ratio $\frac{P_1}{P_2} = \frac{400}{500} = \frac{4}{5}$.

9. Question 5: Heat Engine Efficiency

The efficiency ($\eta$) of a heat engine is given by $\eta = \frac{\text{Work done}}{\text{Heat input}} = \frac{W}{Q_1}$. From the diagram, the source temperature is $327^\circ C = 600$ K and the sink temperature is $27^\circ C = 300$ K. The maximum possible efficiency (Carnot efficiency) is $\eta = 1 - \frac{T_2}{T_1}= 1 - \frac{300}{600} = 1 - \frac{1}{2} = \frac{1}{2} = 0.5 = 50%$.