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A particle moves with constant acceleration along a straight line streaing from rest. The percentage increase in its displacement during the 4th second compared to that in the 3rd second isA. `33%`B. `40%`C. `66%`D. `77%` |
Answer» Correct Answer - B We know that `s_(nth)=u+(1)/(2)a(2n-1)` `s_(3rd)=0+(1)/(2)a(2xx3-1)=(5)/(2)a " " ("for " n= 3s)` `s_(4rd)=0+(1)/(2)a(2xx4-1)=(7)/(2)a" " ("for " n=4s)` So, the percentage increase `=(s_(4th)-s_(3rd))/(s_(3rd))xx100` `=((7)/(2)a-(5)/(2)a)/((5)/(2)a)xx100=((2a)/(2))/((5)/(2)a)xx100=2xx20=40%` |
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