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A heavy uniform rod of weight w is balanced with the help of two strings joined at its left and right ends. If the string at the right end is snapped, then the ratio of tension in the left string in the first and second cases will be:

Option: 1

2 : 1


Option: 2

1 : 2


Option: 3

4 : 1


Option: 4

1 : 4


Answers (1)

best_answer

In Case 1, the tension in the left string is:

                    T = \frac{W}{2}

For Case 2:  

T = W(\frac{l}{2})

\alpha (\frac{1}{3} ml^2) = W(\frac{l}{2})

\alpha = \frac{3}{2} (\frac{W}{ml})

a = \frac{l}{2}\alpha = \frac{3}{2} (\frac{W}{ml})(\frac{l}{2}) = \frac{3}{4} \frac{W}{m}

ma = \frac{3}{4} W

W - T' = ma

T' = W - \frac{3}{4}W = \frac{W}{4}

\frac{T}{T'}= \frac{\frac{W}{2}}{\frac{W}{4}} = \frac{2}{1}

Posted by

Suraj Bhandari

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