A particle starting from point A on a circular path- travelling upto point B with constant speed v1- then upto point C with constant speed v2 and finally upto point A with constant speed v3 as shown below- If particle covers each part of the path in equal intervals of time- then- - - - - - A - - - - - - - - - - - - v1 - - B - - - - v2 - - C - - - - - - v3 - - A - - - - - - - - - - - - - - - - -

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Relative Position

A particle st...

Question

A particle starting from point A on a circular path, travelling upto point B with constant speed v₁, then upto point C with constant speed v₂ and finally upto point A with constant speed v₃ as shown below. If particle covers each part of the path in equal intervals of time, then

एक कण वृत्तीय पथ पर बिंदु A से गति करना प्रारम्भ करता है, यह नीचे दर्शाए अनुसार नियत चाल v₁ से बिंदु B तक, फिर नियत चाल v₂ से बिंदु C तक तथा अन्त में नियत चाल v₃ से बिंदु A तक गति करता है। यदि कण पथ का प्रत्येक भाग समान समयान्तराल में तय करता है, तब

v₃ > v₁

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v₂ > v₃

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v₁ > v₂

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v₁ > v₃

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Solution

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The correct statement with regard to $H_{2}^{+}$ & $H_{2}^{-}$ is

40 m/s

50 m

N O

H_{2}

P_{H_{2}}

(mm of Hg)	359	300	152
$- \frac{P_{H_{2}}}{d t}$	1.50	1.03	0.25

P_{N O}

$P_{H_{2}}$ (mm of Hg)	289	205	147
$- \frac{P_{H_{2}}}{d t}$	1.60	1.10	0.79

H_{2}

N O

H_{1}, H_{2}

2 H . M^{'} . s

a

b

\frac{H_{1} + H_{2}}{H_{1}, H_{2}} =

Assertion : For drying $H_{2} S$ , Conc. $H_{2} S O_{4}$ cannot be used.

Reason: Moist $H_{2} S$ is oxidized by Conc. $H_{2} S O_{4}$

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