Solution
Given : $$p+\frac{1}{p}=\sqrt{10}$$
Squaring both sides, we get :
=> $$(p+\frac{1}{p})^2=(\sqrt{10})^2$$
=> $$p^2+\frac{1}{p^2}+2(p)(\frac{1}{p})=10$$
=> $$p^2+\frac{1}{p^2}=10-2=8$$
Again squaring both sides, we get :
=> $$(p^2+\frac{1}{p^2})^2=(8)^2$$
=> $$p^4+\frac{1}{p^4}+2(p^2)(\frac{1}{p^2})=64$$
=> $$p^4+\frac{1}{p^4}=64-2=62$$
=> Ans - (C)
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