A stone of mass 10 kg having relative density 4 is lying at the bottom of a lake. Find the work done in lifting the stone through a height of 20 m inside the lake
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Answered by
10
M=10. d=4. density= mass/volume 10/4=2•5kgm`3. h=20. p.c = mgh 10×9•5×20=1960joule
Answered by
66
Answer:1500 Joules
Explanation:
Work done=mgh
Apparent weight=Original weight - weight of water displaced
Volume of water displaced = volume of the stone
Which is 1/4000=0.0025m3
Weight of water displaced = volume of water displaced × density × weight of stone
=0.0025×1000×10=25N
So Apparent weight=10×10-25=75N
Therefore, work=75×20=1500 Joules
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