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Standard XII

Chemistry

Heterogeneous Catalysis

The shape of ...

Question

The shape of the string is drawn at $t = 0$ and the area of the pulse enclosed by the string and the x-axis is measured. It will be equal to

2 c m^{2}

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2.5 c m^{2}

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4 c m^{2}

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5 c m^{2}

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Solution

The correct option is A $2 c m^{2}$
$t = 0$

Along x direction

$\frac{x}{4} + 1 = - x + 1 \frac{x}{4} + x = 0 x = 0$

Linear mass density of the string $= 50 g / m$

Tension $= 50 N$

$50 + 50 = 100 N N = \frac{K g / m^{2}}{g / m} \frac{500 N}{50 g / m} = 2 c m^{2} N = m^{2}$

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The shape of the string is drawn at t=0 and the area of the pulse enclosed by the string and the x-axis is measured. It will be equal to

The correct option is A 2 cm2t=0 Along x direction x4+1=−x+1x4+x=0x=0 Linear mass density of the string =50g/m Tension=50N 50+50=100NN=Kg/m2g/m500N50g/m=2cm2N=m2

The shape of the string is drawn at $t = 0$ and the area of the pulse enclosed by the string and the x-axis is measured. It will be equal to

The correct option is A $2 c m^{2}$
$t = 0$

Along x direction

$\frac{x}{4} + 1 = - x + 1 \frac{x}{4} + x = 0 x = 0$

Linear mass density of the string $= 50 g / m$

Tension $= 50 N$

$50 + 50 = 100 N N = \frac{K g / m^{2}}{g / m} \frac{500 N}{50 g / m} = 2 c m^{2} N = m^{2}$