Laws of Motion

  • Question 541
    CBSEENPH11020714

    A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance  and the time are e1 and e2 respectively, the percentage error in the estimation of g is

    • e1 - e2

    • e1 + 2e2

    • e1+ e2

    • e1 - 2e2

    Solution

    B.

    e1 + 2e2

    straight h space equals 1 half gt squared

straight g space equals space straight h over straight t squared
therefore space log space equals space log space straight h space minus 2 space log space straight t
open parentheses fraction numerator increment straight g over denominator straight g end fraction straight x 100 close parentheses subscript max space equals space open parentheses fraction numerator increment straight h over denominator straight h end fraction straight x 100 close parentheses plus 2 open parentheses fraction numerator increment straight t over denominator straight t end fraction straight x 100 close parentheses
straight e subscript 1 plus 2 straight e subscript 2
    Question 542
    CBSEENPH11020716

    An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are 1 kg first part moving with a velocity of 12 ms-1 and 2 kg second part moving with a velocity of 8 ms-1.If the third part flies off with a velocity of 4 ms-,its mass would be

    • 5 kg 

    • 7 kg

    • 17 kg

    • 3 kg

    Solution

    A.

    5 kg 

    apply the law of conservation of linear momentum. 
    momentum of first part = 1 x 12 = 12 kg ms-1
    Momentum of the second part  = 2 x 8 = 16 kg ms-1 '
    Resultant monmentum
    = [(12)2 +(16)2]1/2 = 20 kg ms-1
    The third part should also have the same momentum.


    Let the mass of third part be M, then 
    4 x M = 20
    M = 5 kg
    Question 543
    CBSEENPH11020722

    The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is 

    • 30 ms-2 downward

    • 4 ms-2  upwards

    •  4 ms-2  downwards

    • 14 ms-2  upwards

    Solution

    B.

    4 ms-2  upwards

    Apparent weight > actual weight, then the lift is accelerating upward.
    Lift is accelerating upward at the rate of a 
    Hence, equation of motion is written as
    R - mg = ma
    28000-20000 = 2000a
    a = 8000/2000 = 4 ms-2 upwards.

    Question 544
    CBSEENPH11020736

    A body under the action of a force,straight F with rightwards arrow on top space equals space 6 space bold i with bold hat on top space minus space 8 space bold j with bold hat on top space plus 10 space bold k with hat on top acquires an acceleration of 1 ms-2. The mass of this body must be

    • 2 square root of 10 space kg
    • 20 kg

    • 10 kg

    • 10 square root of 2 space kg

    Solution

    D.

    10 square root of 2 space kg

    According to Newton's second law of motion, force = mass x acceleration.
    Here, 
    straight F with rightwards arrow on top equals space 6 space bold i with bold hat on top space minus 8 space bold j with bold hat on top space plus 10 space bold k with hat on top
vertical line straight F vertical line space equals space square root of 36 plus 64 plus 100 end root

space equals space 10 square root of 2 space straight N
straight a space equals space 1 space ms to the power of negative 2 end exponent
straight m space equals space fraction numerator 10 square root of 2 over denominator 1 end fraction space equals 10 space square root of 2 straight g

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