CONCEPT: Tension in String: The reaction force acting on a tight string when it is holding something is called tension in the string. Pully: Pully is the type of simple machine used to lift any weight by applying lesser force. When not mentioned, the mass of pully and string is considered to be negligible. Freebody Diagram: The free body diagram of an object represents all forces acting with their directions. When two objects are connected to the pulley as shown in the figure then we assume the body with more mass is descending with acceleration and the body with lesser mass is getting uplifted with the same acceleration. Acceleration and Tension offered by the massless Pulley and string are the same throughout the string. The weight of an object is the product of its mass and acceleration due to gravity. CALCULATION: Given mA = 5kg, mB = 10 kg, and g = 10 m/sec2 Let T be the tension in the string. Let a, b be the accelerations of both the blocks A and B. Since the mass of block B is more than block A, so block A will move upward, and block B moves downward. The free-body diagram of block A is drawn as, [ alt="F1 Prabhu Madhuri 22.09.2021 D8" src="//storage.googleapis.com/tb-img/production/21/09/F1_Prabhu_Madhuri_22.09.2021_D8.png" style="width: 155px; height: 175px;"> -----(1) By figure 1, ⇒ mAa = T + T - mAg ⇒ mAa = 2T - mAg ⇒ 5a = 2T - 50 -----(1) The free-body diagram of block B is drawn as, [ alt="F1 Prabhu Madhuri 22.09.2021 D9" src="//storage.googleapis.com/tb-img/production/21/09/F1_Prabhu_Madhuri_22.09.2021_D9.png" style="width: 156px; height: 173px;"> -----(2) By figure 2, ⇒ mBb = mBg - T ⇒ 10b = 100 - T ⇒ T = 100 - 10b -----(2) Now, we need to find the relation between the acceleration of mass A and B which can be found by using the concept of constant length of massless string or by using the tension trick. ΣFA = 0 If F and A are in the same direction then take the product as positive or else negative. For mass A, FA = 2Ta (As A is moving upward by our assumption) For mass B, FA = -Tb, (As B is moving downward by our assumption) ⇒ 2Ta + T(-b) = 0 ⇒ b = 2a ....(3) By, 2 and 3, ⇒ 2T = 200 - 20 (2a) ⇒ 2T = 200 - 40a -----(4) By equation 1 and equation 4, ⇒ 5a = 200 - 40a - 50 ⇒ 45a = 150 ⇒ a = 10/3 m/sec2 ⇒ b = 2(10/3) ⇒ b = 20/3 m/sec2 Hence, option 4 is correct.