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Chapter 1: Problem 64

If \(\mathrm{x}\) meter is a unit of length then area of $1 \mathrm{~m}^{2}=\ldots \ldots \ldots \ldots$ (a) \(x\) (b) \(\mathrm{x}^{2}\) (c) \(x^{-2}\) (d) \(x^{-1}\)

Short Answer

Expert verified
1 square meter (m^2) = \(x^2\) (Option b).

Step by step solution

01

Area of a Square

To calculate the area of a square, we simply multiply the length of one side by itself. In this case, since our side length is 1 meter, we will have an area of \(1m * 1m\).
02

Express the Area in Terms of x

Since we know that x meters is equivalent to 1 meter, we can rewrite the 1 meter sides of the square in terms of x: \(x * x\).
03

Simplify the Expression

Now, multiplying x by itself gives us \(x^2\).
04

Choose the Correct Option

We now see that the area of 1 square meter, in terms of x, is given by the expression \(x^2\). This corresponds to option (b). So, the correct answer is: 1 square meter (m^2) = \(x^2\) (Option b).

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Most popular questions from this chapter

Dimensional formula of latent heat is ........ (a) \(\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\) (b) \(\mathrm{M}^{2} \mathrm{~L}^{0} \mathrm{~T}^{-2}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\) (d) \(\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-1}\)

\( 1 \mathrm{rad}=\ldots \ldots \ldots \ldots\) (a) \(180^{\circ}\) (b) \(3.14^{\circ}\) (c) \((180 / \pi)^{\circ}\) (d) \((\pi / 180)^{\circ}\)

Dimensional formula for torque is (a) \(\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-3}\) (b) \(\mathrm{M}^{2} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\)

If value of gravitational constant in MKS is $6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}\( then value of \)\mathrm{G}$ in \(\mathrm{CGS}=\ldots \ldots \ldots \ldots . .\left[\left(\right.\right.\) dyne \(\left.\left.-\mathrm{cm}^{2}\right) /\left(\mathrm{gm}^{2}\right)\right]\) (a) \(6.67 \times 10^{-9}\) (b) \(6.67 \times 10^{-7}\) (c) \(6.67 \times 10^{-8}\) (d) \(6.67 \times 10^{-5}\)

If $\mathrm{P}=\left[\left(\mathrm{A}^{2} \mathrm{~B}\right) /\left(\mathrm{C}^{3}\right)\right]\( where percentage error in \)\mathrm{A}, \mathrm{B}\( and \)\mathrm{C}$ are respectively \(\pm 2 \% \pm 3 \%\) and \(\pm 5 \%\) then total percentage error in measurement of \(\mathrm{p}\) (a) \(18 \%\) (b) \(14 \%\) (c) \(21 \%\) (d) \(12 \%\)
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