CBSE Class 9 Answered
A WOODEN BLOCK OF MASS 8KG IS TIED TO A STRING ATTACHED TO THE BOTTOM OF THE TANK EQUILIBRIUM THE BLOCK IS COMPLETELY IMMERSED IN WATER .IF THE RELATIVE DENSITY OF WOOD IS 0.8 AND G =10M/S2 THEN WHAT IS THE VAUE OF TENSION IN THE STRING .
Asked by M.s.srinidhi21 | 19 Oct, 2014, 05:09: PM
Expert Answer
Given:
Mass of the wooden block, m= 8kg
Relative density of wood = 0.8
We know that:
The upthrust experienced by the wooden block upward will be balanced by the weight of the wooden block acting downwards + the tension on the string
i.e. Upthrust = Weight + Tension
Therefore the tension on the string = Upthrust - Weight
The upthrust experienced by the wooden block = Volume of water displaced by the block × Density of water × g
Tension on the string = Upthrust - Weight
= 100 N - (8 ×10) N
=20 N
Therefore Tension on the sring = 20N
Answered by Jyothi Nair | 20 Oct, 2014, 10:54: AM
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