Question

A block of mass 10 kg is suspended in air with the help of a light string .Find out the value of tension(T) developed in the string.(g=10m/s^2)​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

m = 10 kg.

g = 10 m/s².

$\huge{\bold{\underline{Explanation:-}}}$

As there is no external force acting on the block.

Therefore, The Tension in the String will be equal to its weight.

$\large{\therefore{T = W} \: \: \: (W = mg)}$

$\large{ \to T = m \times g}$

$\large{ \to T = 10 \times 10}$

$\large{ \to T = 100N}$

$\huge{\boxed{\boxed{T = 100N}}}$

So,the Tension developed in the string is 100N.

Answer 2

Answer:

as the system in equilibrium so exhilaration will be equal to 0 trnsion equal to 10 G which is basically 100 n