JEE physics

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Question
CBSEENPH11020370

A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is 11 kms−1, the escape velocity from the

  • 1.1 kms−1

  • 11 kms−1

  • 110 kms−1

  • 0.11 kms−1

Solution

C.

110 kms−1

straight V subscript esc space equals space square root of fraction numerator 2 GM over denominator straight R end fraction end root space equals space square root of fraction numerator 2 straight G space straight x space 10 space straight M over denominator straight R divided by 10 end fraction end root space
space equals space 10 space straight x space 11 space equals space 110 space km divided by straight s

Sponsor Area

Question
CBSEENPH11020372

A thin rod of length ‘L’ is lying along the x-axis with its ends at x = 0 and x = L. Its linear density (mass/length) varies with x as,straight k space open parentheses straight x over straight L close parentheses to the power of straight n where n can be zero or any positive number. If the position xCM of the centre of mass of the rod is plotted against ‘n’, which of the following graphs best approximates the dependence of xCM on n? 

Solution

A.

space straight x subscript cm space equals space fraction numerator integral dmx over denominator integral dm end fraction
fraction numerator integral begin display style λdx end style begin display style. end style begin display style straight x end style over denominator integral begin display style dm end style end fraction
space equals space fraction numerator integral begin display style straight k end style begin display style begin display style open parentheses straight x over straight L close parentheses end style squared end style begin display style. end style begin display style xdx end style over denominator straight k open parentheses begin display style straight x over straight L end style close parentheses to the power of straight n space dx end fraction
space equals space open square brackets fraction numerator begin display style fraction numerator kx to the power of straight n plus 2 end exponent over denominator left parenthesis straight n plus 2 right parenthesis straight L to the power of straight n end fraction end style over denominator begin display style fraction numerator kx to the power of straight n plus 1 end exponent over denominator left parenthesis straight n plus 1 right parenthesis straight L to the power of straight n end fraction end style end fraction close square brackets subscript 0 superscript straight L
space equals space open square brackets fraction numerator straight x left parenthesis straight n plus 1 right parenthesis over denominator straight n plus 2 end fraction close square brackets subscript 0 superscript straight L
straight x subscript cm space equals space straight L over 2 comma fraction numerator 2 straight L over denominator 3 end fraction comma fraction numerator 4 straight L over denominator 5 end fraction comma fraction numerator 5 straight L over denominator 6 end fraction comma.....

Question
CBSEENPH11020374

The dimension of magnetic field in M, L, T and C (Coulomb) is given as

  • MLT−1C−1

  • MT2C−2

  • MT−1C−1

  • MT−2C−1

Solution

C.

MT−1C−1

F = qvB
B = F/qv
= MC−1 T−1

Question
CBSEENPH11020375

Consider a uniform square plate of side ‘a’ and mass ‘m’. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

  • (5/6)ma2

  • (1/12)ma2

  • (7/12)ma2

  • (2/3)ma2

Solution

D.

(2/3)ma2

straight I space equals straight I subscript cm space plus space straight m open parentheses fraction numerator straight a square root of 2 over denominator 2 end fraction close parentheses squared space equals space ma squared over 6 space plus ma squared over 2
space equals space 2 over 3 ma squared

Question
CBSEENPH11020376

A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms−1.The magnitude of its momentum is recorded as

  • 17.6 kg ms−1 

  • 17.565 kg ms−1

  • 17.56 kg ms−1

  • 17.57 kg ms−1

Solution

A.

17.6 kg ms−1 

P = mv = 3.513 × 5.00 ≈ 17.6