A) increase
B) decrease
C) remain unaffected
D) become infinite
Correct Answer: B
Solution :
\[T=2\pi \sqrt{\frac{l}{g}}\] When train accelerates, the effective value of g changes to g' given by\[(g')=\sqrt{{{a}^{2}}+{{g}^{2}}}\] \[\therefore \] \[T'=\sqrt{\frac{l}{g'}}\] As \[g'>g,T'<T\]You need to login to perform this action.
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