Practice India questions with answers and explanations.
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KE=½mv²=128.0 J.
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V=IR=20 V.
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R=13 Ω.
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mgh=70 J.
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s=vt=48 m.
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KE=½mv²=364.5 J.
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V=IR=10 V.
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R=11 Ω.
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mgh=40 J.
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Four sigma bonds give tetrahedral sp³ hybridisation.
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For fixed gas at constant temperature, PV is constant.
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s=vt=15 m.
