A pendulum has time period T in air. When it is made to oscillate in water, it acquired a time period T'=2T. The specific gravity of the pendulum bob is equal to :

The effective acceleration of a bob in water =g'​=g(1−dD), where d and D are the density of water and the ​ bob respectively. Dd= specific gravity of the bob. ​ Since the period of oscillation of the bob in air and water are  given as ​T=2πℓg and T'=2πℓg'​∴ T/T'=g'g=g(1−d/D)g=1−dD=1−1s​ Putting T/T'=1/2, we obtain ​⇒12=1−1s​⇒1s=12⇒s=2