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$root is the product of n I fractions.) The value of n is (A) 81 (11) 64 (k) 16 (i) 256 (1)) 100$

root is the product of $n$ I fractions.) The value of $n$ is $\newline$ (A) $81$ $\newline$ ( $11$ ) $64$ $\newline$ (k) $16$ $\newline$ (i) $256$ $\newline$ ( $1$ )) $100$

Full solution

Q. root is the product of

n

I fractions.) The value of

n

\newline

(A)

81

\newline

(

11

)

64

\newline

(k)

16

\newline

(i)

256

\newline

(

1

))

100

Understand the question: First, let's understand the question. We're looking for a number $n$ , that when we take its square root, we get the product of $n$ one over $n$ fractions. So basically, we're looking for a number that when multiplied by itself gives us $n$ .
Check option (A) $81$ : Let's try the options given one by one. Starting with option (A) $81$ . If we take the square root of $81$ , we get $9$ . Now, $9 \times \frac{1}{9}$ is $1$ , not $81$ . So, this ain't it.
Check option (B) $64$ : Next, let's try option (B) $64$ . The square root of $64$ is $8$ . And $8$ times $1/8$ is $1$ , not $64$ . So, this ain't right either.
Check option (C) $16$ : Now, let's check option (C) $16$ . The square root of $16$ is $4$ . And $4$ times $1/4$ is $1$ , not $16$ . So, this one's wrong too.
Check option (D) $256$ : Moving on to option (D) $256$ . The square root of $256$ is $16$ . And $16 \times \frac{1}{16}$ is $1$ , not $256$ . So, this one's incorrect as well.
Check option (E) $100$ : Finally, let's check option (E) $100$ . The square root of $100$ is $10$ . And $10$ times $1/10$ is $1$ , which is what we're looking for! So, the value of $n$ is $100$ .