A simple pendulum has a time period T1. When its point of suspension is moved vertically upwards according as y = kt², where y is vertical distance covered and k = 1 ms-2, its time period becomes T2. Then, \(\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}\) is (g = 10 ms-2
A. \(\frac { 5 }{ 6 }\)
B. \(\frac { 11 }{ 10 }\)
C. \(\frac { 6 }{ 5 }\)
D. \(\frac { 5 }{ 4 }\)
Answer:C
A simple pendulum has a time period T1. When its point of suspension is moved vertically upwards according as y = kt², where y is vertical distance covered and k = 1 ms-2, its time period becomes T2. Then, \(\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}\) is (g = 10 ms-2
A. \(\frac { 5 }{ 6 }\) B. \(\frac { 11 }{ 10 }\) C. \(\frac { 6 }{ 5 }\) D. \(\frac { 5 }{ 4 }\)