If the coefficient of x in the expansion of ( | vedclass

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(C) The general term in the expansion of $(x^2 + \frac{k}{x})^5$ is given by $T_{r+1} = {}^5C_r (x^2)^{5-r} (\frac{k}{x})^r$.Simplifying the expressio

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$3$

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(C) The general term in the expansion of $(x^2 + \frac{k}{x})^5$ is given by $T_{r+1} = {}^5C_r (x^2)^{5-r} (\frac{k}{x})^r$.Simplifying the expression,we get $T_{r+1} = {}^5C_r x^{10-2r} k^r x^{-r} = {}^5C_r k^r x^{10-3r}$.For the coefficient of $x$,we set the exponent of $x$ equal to $1$:$10 - 3r = 1$ $\Rightarrow 3r = 9$ $\Rightarrow r = 3$.Substituting $r = 3$ into the coefficient expression:Coefficient $= {}^5C_3 k^3 = 10 k^3$.Given that the coefficient is $270$:$10 k^3 = 270$ $\Rightarrow k^3 = 27$ $\Rightarrow k = 3$.

Column $-I$	Column $-II$
$(a)$ Parturition	$(p)$ Attachment of embryo to endometrium
$(b)$ Gestation	$(q)$ Release of egg from Graafian follicle
$(c)$ Ovulation	$(r)$ Delivery of Foetus from uterus
$(d)$ Implantation	$(s)$ Duration between pregnancy and birth
	$(t)$ Formation of zygote by fusion of egg and sperm

If the coefficient of $x$ in the expansion of $(x^2 + \frac{k}{x})^5$ is $270$,then $k =$

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