Q: One end of a horizontal uniform beam of weight W and length L is hinged on a vertical wall at point O and its other end is supported by a light inextensible rope. The other end

∵ Rod is in equilibrium, net torque on the rod about O will be zero. W l 2 + α W l = T 2 l ⇒ W 2 + α W = T 2 ⇒ T = 2 W α + 0 . 5 . . . 1 Now checking options- (A) Force balance on rod, N v + T 2 = W + α W ⇒ N v = W + α W - W 2 + α W ⇒ N v = W 2 Hence, option (A) is correct. (B) Write horizontal force equation, N H = T 2 ⇒ N H = W α + 0 . 5 If α = 0 . 5 , the

Question 1

A person sitting inside an elevator performs a weighing experiment with an object of mass 50 kg . Suppose that the variation of the height y (in m ) of the elevator, from the ground, with time t (in s

Accepted Answer

y=8+8 2 t T With respect to elevator, variation in weight will be aligned & W = m ( a )_ \ & W = m 2 ^2 ~A aligned Here elevator is performing SHM aligned & W =2 ~m ( 2 ~T )^2 A N \ & W =2 50 ( 2 40 )^2 8 ~N \ & ~W =2 50 1 400 8 ~N \ & ~W = 800 400 ~N =2 ~N aligned

Question 2

One end of a horizontal uniform beam of weight  W and length  L  is hinged on a vertical wall at point  O and its other end is supported by a light inextensible rope. The other end

Accepted Answer

&#8757;  Rod is in equilibrium, net torque on the rod about  O  will be zero. W l 2 + &#945; W l = T 2 l &#8658; W 2 + &#945; W = T 2 &#8658; T = 2 W &#945; + 0 . 5 &#160; &#160; &#160; . . . 1   Now checking options- (A) Force balance on rod, N v + T 2 = W + &#945; W &#8658; N v = W + &#945; W - W 2 + &#945; W &#8658; N v = W 2 Hence, option (A) is correct. (B) Write horizontal force equation, N H = T 2 &#8658; N H = W &#945; + 0 . 5 If  &#945; = 0 . 5 , the

Question 3

A student skates up a ramp that makes an angle 30 ° with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed v 0 and wants to turn around over a semicircular pa

Accepted Answer

Speed at  x and  z is equal and also maximum in circular track. From energy conversation, 1 2 m v 0 2 = m g h + 1 2 m v 2       .......(1) at y , &#160; mg sin 30 &#176; = m v 2 R &#8658; v 2 = R g 2 &#8756;  From equation (1) v 0 2 - 2 g h = R g 2

Question 4

Put a uniform meter scale horizontally on your extended index fingers with the left one at 0 . 00 &#160; cm and the right one at 90 . 00 &#160; cm . When you attempt to move both the fingers slowly to

Accepted Answer

5 x ' = 40 x ' = 40 5 ∴ 4 x ' = 32 5 x ' ' = 32 x ' ' = 32 5 ∴ 4 x ' ' = 4 5 × 32 = x R x R = 25 . 6

Laws of Motion — Previous Year Questions & Solutions