A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass'm'is suspended from the rod at 160 cm mark as shown in the figure. Find the value of 'm' such that the rod is in equilibrium. (g= 10 m/s2)
1)1/2 kg
2) 1/3 kg
3)1/6 kg
4) 1/12 kg
Answers
Answer:
answer is as follows
Option 4) 1/12
By balancing torque, 2g x 20 = 0.5g x 60 + mg x 120 m = 0.5/6 kg = 1/12 kg
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Answer: (4) 1/12 kg
Given:
i) Length of a uniform rod = 200 cm
ii) Mass of a uniform rod = 500 g = 0.5 kg
iii) The uniform rod is balanced on a wedge placed at 40 cm mark
iv) A mass of 2 kg is suspended from the rod at 20 cm
v) Another unknown mass 'm' is suspended from the rod at 160 cm mark
To find:
The value of 'm' such that the rod is in equilibrium
Solution:
We know that for a body to be in state of equilibrium, the following 3 conditions should be satisfied:
a) Fnet along any direction on the body = 0
b) Tnet about any point on the body = 0
where
F: Force
T: Torque
Using the fact that Tnet about any point on the body = 0, when the body is in equilibrium, and calculating the torque of all the forces acting on the body about the 40 cm mark, we get
(Sign convention: Anticlockwise Torque: +ve
Clockwise Torque: -ve)
Torque due to weight of the rod at 100 cm mark (T1) = + (0.5*10)(100 - 40)
= + (5)(60)
= + 300 Ncm
Torque due to mass of 2 kg at 20 cm mark (T2) = - (2*10)(40 - 20)
= - (20)(20)
= - 400 Ncm
Torque due to mass 'm' kg at 160 cm mark (T3) = + (m*10)(160 - 40)
= + (m*10)(120)
= + 1200*m Ncm
Torque due to normal due to wedge at 40 cm mark (T4) = (N)(40 - 40)
= 0
Since the body is in equilibrium,
Resultant T = 0
=> T1 + T2 + T3 + T4 = 0
=> +300 + (-400) + 1200*m + 0 = 0
=> -100 + 1200*m = 0
=> 1200*m = 100
=> m = 1/12 kg
The value of 'm' = 1/12 kg
Hence, option (4) is correct
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