This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Question 10: Period of a simple pendulum Step 1: Recall the formula for the period of a simple pendulum. The period (T) of a simple pendulum is given by: T = 2 sqrt((L)/(g)) Where L is the length of the pendulum and g is the acceleration due to gravity. Step 2: Analyze the relationship between the period and length. From the formula, the period T is directly proportional to the square root of the length L: T sqrt(L) Step 3: Determine the change in period when the length is increased by a factor of 4. Let the original length be L_1 and the original period be T_1. T_1 = C sqrt(L_1) Where C = 2/sqrt(g) is a constant. If the length is increased by a factor of 4, the new length L_2 = 4L_1. The new period T_2 will be: T_2 = C sqrt(L_2) = C sqrt(4L_1) T_2 = C × sqrt(4) × sqrt(L_1) T_2 = 2 × (C sqrt(L_1)) T_2 = 2 T_1 The new period is twice the original period, meaning it is increased by a factor of 2. The period will be c) increased by 2. --- Question 11: Value of external resistance R Step 1: Identify the given information. Internal resistance of the cell, r = 2 . The terminal voltage (V) is reduced to (3)/(5) of its nominal value (EMF, E). So, V = (3)/(5) E. Step 2: Recall the formula for terminal voltage. The terminal voltage (V) across an external resistance (R) connected to a cell with EMF (E) and internal resistance (r) is given by: V = E ((R)/(R + r)) Step 3: Substitute the given values into the formula. We have V = (3)/(5) E and r = 2 . (3)/(5) E = E ((R)/(R + 2 )) Step 4: Solve for R. Divide both sides by E (assuming E ≠ 0): (3)/(5) = (R)/(R + 2) Cross-multiply: 3(R + 2) = 5R 3R + 6 = 5R Subtract 3R from both sides: 6 = 5R - 3R 6 = 2R Divide by 2: R = (6)/(2) R = 3 The value of R is c) 3 .