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Magnitude using Vector: Derivation

Th figure sho...

Question

Th figure shows a block of mass $m$ moving without friction along three tracks with same speed $v$ .

Choose the correct relation

N_{1} > N_{2}

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N_{2} > N_{3}

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N_{3} > N_{1}

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N_{1} = N_{2} = N_{3}

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Solution

The correct options are
A $N_{1} > N_{2}$
C $N_{3} > N_{1}$
Case 1
$N_{1} = m g$
Case 2

$m g - N_{2} = \frac{m v^{2}}{r}$

Case 3 $N_{3} = \frac{m v^{2}}{r} + m g$
$∴ N_{1} > N_{2}; N_{3} > N_{1}$

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Similar questions

m

v

C_{1}, C_{2}, C_{3}

n_{1}, n_{2}, n_{3}

{(\frac{n_{1} C_{1}^{2} + n_{2} C_{2}^{2} + n_{3} C_{1}^{3} + \dots}{n_{1} + n_{2} + n_{3} + \dots})}^{\frac{1}{2}}

\frac{(n_{1} C_{1}^{2} + n_{2} C_{2}^{2} + n_{3} C_{1}^{3} + \dots)^{\frac{1}{2}}}{n_{1} + n_{2} + n_{3} + \dots}

\frac{(n_{1} C_{1}^{2})^{\frac{1}{2}}}{n_{1}} + \frac{(n_{2} C_{2}^{2})^{\frac{1}{2}}}{n_{2}} + \frac{(n_{3} C_{1}^{3})^{\frac{1}{2}}}{n_{3}} + \dots

{[\frac{(n_{1} C_{1}^{2} + n_{2} C_{2}^{2} + n_{3} C_{1}^{3} + \dots]^{2}}{(n_{1} + n_{2} + n_{3} + \dots)}]}^{\frac{1}{2}}

If $C_{1}, C_{2}, C_{3}$ .........represent the speeds of $n_{1}, n_{2}, n_{3}$ ......... molecules, then the root mean square speed is _____.

800 m/s

799 - 800 m/s

N_{1}

899 - 900 m/s

N_{2}

999 - 1000 m/s

N_{3}

5 < n_{1} < n_{2} < n_{3} < n_{4}

n_{1} + n_{2} + n_{3} + n_{4} = 35

(n_{1}, n_{2}, n_{3}, n_{4})

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