Q.

In the relation v=π8Pr4nl,  where the letters have their usual meanings, the dimensions of v are

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a

M0L3T0

b

M0L3T−1

c

M0L−3T−1

d

M1L3T0

answer is B.

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Detailed Solution

V=π8Pr4nl=(ML−1T−2)L4(ML−1T−1)L=M0L3T−1
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In the relation v=π8Pr4nl,  where the letters have their usual meanings, the dimensions of v are