Question

25.
A uniform rod has a weight of 40 N and a length of 1 m. It is hinged to a wall (at the left end), and held in a
horizontal position by a vertical massless string (at the right end). What is the magnitude of the torque exerted
by the string about a horizontal axis which passes through the hinge and is perpendicular to the rod?
String
Hinge
Rod
Wall
(2) 10 Nm
(1) 5 Nm
(4) 40 Nm
(3) 20 Nm
L
l
ongth of 1 m
It is hinged to a wall at the left end), and held in a​

Answer 1

Answer:

Perior Concept=Weight act at the center of of body

Torque= Force × perpendicular distance

Given that: weight(w) =40N

Distance from the center of rod to hinged point

=.5m

Torque= W×L/2

=40×1/2

=20 N ANS.

Answer 2

The torque exerted by the string is 20 Nm.

(3) is correct option.

Explanation:

Given that,

Weight of rod= 40 N

Length of rod = 1 m

Rod is at rest. it is in rationally equilibrium

So net torque about point A is zero.

We need to calculate the torque exerted by the string

Using formula of torque

$\tau_{mg}-\tau_{s}=0$

$\tau_{mg}=\tau_{s}$

$mg\times\dfrac{l}{2}=\tau_{s}$

Where, $\tau_{s}$ is the torque exerted by the string

Put the value into the formula

$\tau_{s}=40\times\dfrac{1}{2}$

$\tau_{s}=20\ Nm$

Hence, The torque exerted by the string is 20 Nm.

Learn more :

Topic : Torque

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