Title 1. Low of motion (pass.Q ) E.pdf ✅

on horizontal smooth surface and a horizontal force F is applied as shown. Points P and Q are in same vertical line and two strings are of equal length m  F P Q Q.16 Minimum force F required to make both string tight is - (A) Mg tan (B) Mg cot  (C) (m + M)g tan (D) (m + M)g cot  Sol. [C] For minimum force F lower string just remain tight i.e. T2  0  T1 cos = mg and T1 sin = ma  a = g tan  F = (m + M)g tan m  a mg T1 T2 ma Q.17 If the lower string is just tight then tension in the upper string is - (A) sin  mg (B) cos mg (C) mg sin (D) mg cos Sol. [B] For minimum force F lower string just remain tight i.e. T2  0  T1 cos = mg and T1 sin = ma  a = g tan  F = (m + M)g tan m  a mg T1 T2 ma Q.18 For what value of force F tension in upper string is twice than tension in lower string. (A) 3(m + M)g tan (B) 2(m + M)g tan (C) 4(m + M)g tan (D) None of these Sol. [A] T1 cos = mg + T2 cos  (T1–T2) cos = mg ..............(i) also (T1+T2) sin = ma ..............(ii) from (i) and (ii)          1 2 1 2 T – T T T tan = g a But as T1 = 2T2 given  a = 3g tan Passage # 7 (Ques. 19to 21) A block of mass m = 2 kg kept on a wedge of mass M = 9 kg and a horizontal force of 210N is applied on the wedge as shown. All the contacts are smooth and use g = 10m/s2 . If initially when the force is applied m block is at the bottom of wedge and base length of wedge is 10 m. M 45o 10m F=210N m Q.19 Acceleration of wedge when mass m is moving on it - (A) 11 210 m/s2 (B) 2 10 m/s2 (C)20 m/s2 (D) 10 2 m/s2 Sol. [C] mg m  a2 N ma1 M a1 Let a2 be acceleration of m w.r.t M Then, ma1 cos – mg sin ma2 2 a1 – 2 g = a2 a1 – g = a2 × 2 ......(i) N = ma1 sin + mg cos ..............(ii)