i) Linear equilibrium mg = N2 + µ1N1 From figure µ2N2 = N1 mg = N2 + µ1µ2N2 (µ1µ2 +1 ) N2 = mg ∴ Option c correct From figure N1 tanθ=N2 ii) From rotational equilibrium Let l be the length of ladder Clock wise motion torque = mg (l/2) cosθ Anti-clock wise motion torque = N1lsinθ Option d correct Answer:(c, d)

For equilibrium condition f = m1gsinθ + m2gsinθ …(i) friction f = µN = μm2gcosθ ..(ii) µm2gcosθ = m1gsinθ + m2gsinθ µm2g =( m1g + m2g) tanθ 0.3 × 2 = (1+2) tanθ tanθ = 0.2 If angle less or equal to 11.5 Masses will not slip List I P ,θ = 5°< 11.5° List I Q ,θ = 10°< 11.5° equilibrium conditions given in (i) hold good Thus P → 2, Q →2 List I R. θ = 15° List I S. θ = 20° Equilibrium condition given in (i) not followed masses will slip And friction will be as per (ii) Thus R → 3, S →3 P → 2, Q →2, R → 3, S →3 Answer: d Answer:(d)

Let level of water in jar be h, and be at height H When lift was not moving velocity of water ejected from hole Time taken for water to reach floor, vertical velocity initial is zero Horizontal distance So long resultant g ≠ 0 P. Lift is accelerating vertically up. Resultant g’ > g, so d is intendant of g , d = 1.2 m Q. Lift is accelerating vertically down with an acceleration less than the gravitational acceleration. Resultant g’ < g, so d is intendant of g , d = 1.2 m R. Lift is moving vertically up with constant speed. Resultant g’ =g, so d is intendant of g , d = 1.2 m S. Lift is falling freely. Resultant g’ =0, so no water will come out of jar P, Q, R → 1 and S → 4 Answer:(c)

Linear equilibrium Nsin30 + N = mg And fr= Ncos30 Rotational equilibrium at point A Option d correct Answer:(d)

Let m be the mass of gas molecules. Plate is moving with velocity v. Speed of gas molecules with respect to leading plate is u+v. gas molecule under goes a collision with plate change in momentum of plate due to one molecule= 2m(u+v). In time ∆t, volume swept by plate of unit area =vΔt If n is the number of molecules per unit volume can be considered as constant as pressure is low. Then number of molecules collide with surface = nv∆t Thus momentum gain by plate per unit area = nvΔt × 2m(u+v). = 2mn (vu+v2)Δt Force on leading face per unit area = PL = change in momentum/Δt = 2mn (vu+v2) For rear face velocity of gas with respect to plate is v-u For rear face force per unit area = PR = 2mn (v2 -uv) Thus force difference per unit area = pressure difference = PL- PR 2mn (vu+v2) - 2mn (v2-uv ) = 2mn (2uv) = 4mn(uv) Thus option [C] correct Force per unit area due to gas = 4mn(uv) thus proportional to v [ option a correct] Net force on plate = F(applied )- F (resistive) Net force on plate = F – 4mn(uv) Here F is constant thus at a particular velocity of plate F = 4mn(uv) and acceleration of plate will be zero [ Option c is wrong and option d correct] Answer:(a, b, d)

Here m2 > m1 .Both the blocks will move with same acceleration 'a' Forces are shown in figure For block m2 m2a = m2g – T … (1) For block m1 m1a = T –fk … (2) To find T we have to eliminate a as we have assumed it (1) + (2) m2a+ m1a = m2g -fk Now fk = µkN and N = m1g ; substituting value of fk = µk m1g ...(3) Answer : (c)