Question

Two bodies of masses 4 kg and 6 kg are attached to the ends of a string passing over a pulley. The 4 kg mass is attached to the table top by another string. The tension in this string T, is equal to (Take g = 10 ms?) т 4Kg. 6Kg A) 20N T B) 251 10 CD 10N​

Pg

Answer 1

Answer:

EXPLANATION.

=> Two bodies of masses 4kg and 6kg are

attached to the ends of a string passing

over a pulley.

=> The 4 kg mass is attached to the table Top

by another string.

To find the tension in the string.

=> For a 6 kg of block

=> T = 6g

=> T = 60 N ........(1)

=> For a 4 kg of block

=> T - T¹ - 4g = 0

=> T = T¹ + 4g

=> 60 = T¹ + 40

=> T¹ = 60 - 40

=> T¹ = 20N

Therefore,

option [ A ] is correct answer.

Answer 2

Explanation:

Question :

Two bodies of masses 4 kg and 6 kg are attached to the ends of a string passing over a pulley. The 4 kg mass is attached to the table top by another string. The tension in this string T, is equal to ( Take g = 10 m/s² )

A) 20 N

B) 25 N

C) 30 N

D) 35 N

Solution :

Apply FBD for 6 kg block ,

$\begin{lgathered}\to \sf 6g=T\\\\\to \sf 6(10)=T\\\\\to \sf T=60\ N\ \; \bigstar\end{lgathered}$

Apply FBD for 4 kg block ,

$\begin{lgathered}\to \sf T=4g+T_1\\\\\to \sf 60-4(10)=T_1\\\\\to \sf 60-40=T_1\\\\\to \sf T_1=20\ N\ \; \bigstar\end{lgathered}$

So , Tension in the string = 20 N

Option A

Cheers :) Best of Luck :)