Question

4. Two masses 7.5 kg and 6.7 kg are attached to the ends of an inextensible string passes
over a frictionless pulley. Its tension when both masses are moving vertically will be
equal to
A. 50 N
C. 65.6 N
B. 70.7 N
D. 45 N​

Answer 1

answer of this question is b part 70.7 N

hope this helps you

Answer 2

The tension when both masses are moving vertically. is 70.7N. So, option 'B' is correct.

Given:-

Mass of first body (m1) = 7.5kg

Mass of second body (m2) = 6.7kg

To Find:-

The tension when both masses are moving vertically.

Solution:-

We can easily find out the tension when both masses are moving vertically by following these simple steps.

As

Mass of first body (m1) = 7.5kg

Mass of second body (m2) = 6.7kg

g = 10m/s²

According to the formula of the Pulley block system tension in the string is given by,

$t = \frac{(2 \times m1 \times m2)g}{m1 + m2}$

By putting the values of m1, m2, and g we get,

$t = \frac{(2 \times 7.5 \times 6.7) \times 10}{7.5 + 6.7}$

$t = \frac{20 \times 50.25}{14.2}$

$t = \frac{1005}{14.2}$

$t = 70.7N$

Hence, The tension when both masses are moving vertically. is 70.7N. So, option 'B' is correct.

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