The following are correct about the human eye and camera except that they

37. The following are correct about the human eye and camera except that they The correct option is C. The human eye can adjust its focal length (accommodation) to focus on objects at different distances. A typical camera lens has a fixed focal length, although modern cameras can use multiple lenses or adjustable lens elements to change the effective focal length. Therefore, having a fixed focal length is the exception when comparing the general characteristics of the human eye and a camera. 38. An object moved with a speed of 150 ms^-1 for 1 minute. Calculate the distance covered by the object. Step 1: Convert the time from minutes to seconds. t = 1 minute × 60 secondsminute = 60 s Step 2: Use the formula for distance, speed, and time. d = v × t d = 150 ms^-1 × 60 s d = 9000 m The distance covered by the object is E. 9000.0 m. 39. A car moving at 10 ms^-1 decelerates at 5 ms^-2 for 1.34 s. Calculate the magnitude of its final velocity. Step 1: Identify the given values. Initial velocity, u = 10 ms^-1 Acceleration (deceleration), a = -5 ms^-2 Time, t = 1.34 s Step 2: Use the first equation of motion to find the final velocity. v = u + at v = 10 ms^-1 + (-5 ms^-2) × 1.34 s v = 10 ms^-1 - 6.7 ms^-1 v = 3.3 ms^-1 The magnitude of its final velocity is A. 3.3 ms^-1. 40. A uniform rod AB of mass 20 kg is pivoted at A. The rod is kept in equilibrium by a rope of tension 1000 N making an angle of 30^ to the rod. If the rope is attached to the rod at a distance of 2 m from the pivot, calculate the length of the rod. [g = 10 ms^-2] Step 1: Calculate the weight of the rod. For a uniform rod, its weight acts at its center of mass, which is at L/2 from the pivot. W = mg = 20 kg × 10 ms^-2 = 200 N Step 2: Calculate the moment due to the weight of the rod about pivot A. This is a clockwise moment. M_W = W × (L)/(2) = 200 N × (L)/(2) = 100L Nm Step 3: Calculate the moment due to the tension in the rope about pivot A. This is an anticlockwise moment. The perpendicular component of the tension is T . M_T = (T ) × d_T M_T = (1000 N × 30^) × 2 m M_T = (1000 N × 0.5) × 2 m M_T = 500 N × 2 m = 1000 Nm Step 4: For equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments. M_W = M_T 100L = 1000 L = (1000)/(100) L = 10 m The length of the rod is E. 10.0 m. 41. A ball of mass 5 kg falls from a height of 120 m. If its potential energy after falling through a distance x is 3.5 kJ, neglecting air resistance, calculate x. [g = 10 ms^-2] Step 1: Convert the potential energy to Joules. PE_f = 3.5 kJ = 3.5 × 1000 J = 3500 J Step 2: Determine the height of the ball after falling a distance x. The initial height is H = 120 m. The height after falling x is h_f = H - x = (120 - x) m. Step 3: Use the potential energy formula PE = mgh to find the final height. PE_f = mgh_f 3500 J = 5 kg × 10 ms^-2 × (120 - x) m 3500 = 50 (120 - x) Step 4: Solve for x. (3500)/(50) = 120 - x 70 = 120 - x x = 120 - 70 x = 50 m The distance x is C. 50 m. 42. A machine with a velocity ratio of 5 was used to move a load of 2000 N with an effort of 500 N. Calculate the efficiency of the machine. Step 1: Calculate the Mechanical Advantage (MA) of the machine. MA = LoadEffort MA = 2000 N500 N MA = 4 Step 2: Calculate the efficiency () of the machine. = (MA)/(VR) × 100\% = (4)/(5) × 100\% = 0.8 × 100\% = 80\% The efficiency of the machine is E. 80\%. 43. A body x of mass 2 kg moving with a velocity of 20 ms^-1 collides with another body y of mass 3 kg moving in the opposite direction. If the bodies stick together and move in the direction of x at a velocity of 6 ms^-1, calculate the velocity of y. Step 1: Identify the given values and set up the conservation of momentum equation. Mass of body x, m_x = 2 kg Initial velocity of body x, u_x = 20 ms^-1 Mass of body y, m_y = 3 kg Initial velocity of body y, u_y = ? (since it's in the opposite direction, it will have a negative sign) Final velocity of combined bodies, v = 6 ms^-1 (in the direction of x) According to the conservation of momentum: m_x u_x + m_y u_y = (m_x + m_y) v Step 2: Substitute the known values into the equation. (2 kg)(20 ms^-1) + (3 kg)u_y = (2 kg + 3 kg)(6 ms^-1) 40 + 3u_y = (5)(6) 40 + 3u_y = 30 Step 3: Solve for u_y. 3u_y = 30 - 40 3u_y = -10 u_y = -(10)/(3) ms^-1 u_y ≈ -3.33 ms^-1 The magnitude of the velocity of y is 3.3 ms^-1. The negative sign indicates it was moving in the opposite direction to body x. The velocity of y is E. 3.3 ms^-1. 44. A metal rod of length 2.01 m at 20^ was heated to 30^. Calculate the new length of the rod. [ = 1.7 × 10^-5 K^-1] Step 1: Calculate the change in temperature ( T). T = T_f - T_0 = 30^ - 20^ = 10^ Since a change of 1^ is equal to a change of 1 K, T = 10 K. Step 2: Calculate the change in length ( L) using the linear thermal expansion formula. L = L_0 T L = (2.01 m) × (1.7 × 10^-5 K^-1) × (10 K) L = 2.01 × 1.7 × 10^-4 m L = 0.0003417 m Step 3: Calculate the new length (L_f). L_f = L_0 + L L_f = 2.01 m + 0.0003417 m L_f = 2.0103417 m Rounding to four decimal places, L_f ≈ 2.0103 m. The new length of the rod is C. 2.0103 m. Drop the next question! 📸