NAME FEGITRATION NUMBER terfall. BRAVCH Choose correct answer (put / ) of the following questions. ( \( 1 \times 10 \) ) | A copper rod of lingeh I m and crosu-sectional area \( 0.005 \mathrm{~m}^{2} \) has its ends maintained at \( 190^{\circ} \mathrm{C} \) and \( 0^{\circ} \mathrm{C} \) If the thermal conductonाy of copper is \( 400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \). The rate of heat tramier is (a) 200 Wath (b) 1470 wall (c) 1050 W att (d) 1260 War 2. A stecl wirc of tength 3 m and crons-sectaunal arca \( 2 \times 10^{*} \mathrm{~m}^{2} \) as atretehed by 1.5 mm under a force of 200 N . The Young's modulus of the material. (a) 200 GPa (b) 100 GPa (c) 300 GPa (d) 150 GPa 3 A cricket ball of mans 0.5 hy is moving with a wolocity of \( 10 \mathrm{~m} / \mathrm{s} \) After being hit by a bat. it moves in the opponice dirction with a velocity of -15 m . The change in mernentum is equal ts (a) (b) \( -125 \mathrm{~kg} m / \mathrm{s} \) (c) \( 12.5 \mathrm{~kg} \mathrm{~m} / \mathrm{s} \) (d) \( 625 \mathrm{~kg} \mathrm{~m} / \mathrm{s} \) 4 The value of \( \frac{6}{c t} \) for air is (d) 14, (b) 1.33 , (c) 1 , (d) 0.8 .

Let's solve each question step by step. ### Question 1 A copper rod of length \( L = 1 \, \text{m} \) and cross-sectional area \( A = 0.005 \, \text{m}^2 \) has its ends maintained at \( T_1 = 190^{\circ} \text{C} \) and \( T_2 = 0^{\circ} \text{C} \). The thermal conductivity of copper is \( k = 400 \, \text{W/m} \cdot \text{K} \). The rate of heat transfer \( Q \) can be calculated using Fourier's law of heat conduction: \[ Q = \frac{k \cdot A \cdot (T_1 - T_2)}{L} \] Substituting the values: \[ Q = \frac{400 \, \text{W/m} \cdot \text{K} \cdot 0.005 \, \text{m}^2 \cdot (190 - 0)}{1} \] Calculating \( Q \): \[ Q = \frac{400 \cdot 0.005 \cdot 190}{1} = 380 \, \text{W} \] None of the options match this result, so let's check the calculations again. ### Question 2 A steel wire of length \( L = 3 \, \text{m} \) and cross-sectional area \( A = 2 \times 10^{-6} \, \text{m}^2 \) is stretched by \( \Delta L = 1.5 \, \text{mm} = 0.0015 \, \text{m} \) under a force \( F = 200 \, \text{N} \). Young's modulus \( Y \) is given by: \[ Y = \frac{F \cdot L}{A \cdot \Delta L} \] Substituting the values: \[ Y = \frac{200 \cdot 3}{2 \times 10^{-6} \cdot 0.0015} \] Calculating \( Y \): \[ Y = \frac{600}{3 \times 10^{-9}} = 200 \times 10^{6} \, \text{Pa} = 200 \, \text{GPa} \] So the answer is (a) 200 GPa. ### Question 3 A cricket ball of mass \( m = 0.5 \, \text{kg} \) is moving with a velocity \( v_1 = 10 \, \text{m/s} \) and after being hit, it moves in the opposite direction with a velocity \( v_2 = -15 \, \text{m/s} \). The change in momentum \( \Delta p \) is given by: \[ \Delta p = m \cdot (v_2 - v_1) \] Substituting the values: \[ \Delta p = 0.5 \cdot (-15 - 10) = 0.5 \cdot (-25) = -12.5 \, \text{kg m/s} \] So the answer is (b) \( -12.5 \, \text{kg m/s} \). ### Question 4 The value of \( \frac{6}{c t} \) for air is not provided with specific values for \( c \) and \( t \). However, if we assume \( c \) is the speed of light in air and \( t \) is time, we can analyze the options given. Without specific values, we cannot calculate this directly. However, if we assume standard values for \( c \) and \( t \), we can evaluate the options. Let's summarize the answers: 1. **Question 1:** None of the options match the calculated value of 380 W. 2. **Question 2:** (a) 200 GPa 3. **Question 3:** (b) \( -12.5 \, \text{kg m/s} \) 4. **Question 4:** Cannot determine without specific values for \( c \) and \( t \). If you have any specific values for Question 4 or need further assistance, please let me know!