A mass of 10 kg is suspended from a spring b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $T$ be the tension in the spring balance. The forces acting on the mass $M$ are the tension $T$ along the string,the gravitational force $mg$

Vedclass · Accepted Answer

$20 \ kg-wt$

Vedclass · Answer

(A) Let $T$ be the tension in the spring balance. The forces acting on the mass $M$ are the tension $T$ along the string,the gravitational force $mg$ acting vertically downwards,and the horizontal force $F$ applied by the string.In equilibrium,the vertical component of the tension $T$ must balance the weight of the mass.$T \cos 	heta = mg$Given $m = 10 \ kg$,$	heta = 60^{\circ}$,and $g$ is the acceleration due to gravity.The reading of the spring balance corresponds to the tension $T$ in $kg-wt$.$T = \frac{mg}{\cos 60^{\circ}}$Since $1 \ kg-wt = 1 \ kg 	imes g$,the reading in $kg-wt$ is simply $T/g = m / \cos 60^{\circ}$.$T_{reading} = \frac{10}{\cos 60^{\circ}} = \frac{10}{1/2} = 20 \ kg-wt$.

$A$ mass of $10 \ kg$ is suspended from a spring balance. It is pulled aside by a horizontal string so that it makes an angle of $60^{\circ}$ with the vertical. The new reading of the balance is

Similar Questions

$A$ spring of force constant $k$ is cut into two pieces such that one piece is three times the length of the other. The longer piece will have a force constant of

If a spring is extended to length $l,$ then according to Hooke's law,

The force required to stretch a spring varies with the distance as shown in the figure. If the experiment is performed with the above spring of half length,the line $OA$ will

Vedclass Test Series

Exam Paper Generator

Online Exam Module