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1)

The bob of a simple pendulum  performs SHM with period  T in air and with period T1 in water, Relation  between T and T1 is (neglect friction due to water , density of the material  of the bob  is = 98×103kgm3 , density of water =1 gcc-1)


A) T1=3T

B) T1=2T

C) T1=T

D) T1=T2

Answer:

Option A

Explanation:

 Time period  of simple pendulum in water is given by

                   T=2πlgeff

[ Symbols have their  usual meanings and   geff = acceleration due to gravity in water]

 As we know, geff=g(σρσ)

 [ where, σ = density of bob.ρ = density of water]

 =9.8(98×10310398×103)

   =9.8(98198)=9.8(18×89)

 =9.89

  So, T1=2πl×99.8T1=3T

 [  Time period of simple pendulum in air, 

                T=2πl9.8]