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Physics Test - 18

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Physics Test - 18
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Weekly Quiz Competition
  • Question 1
    1 / -0

    Which of the following is a scalar quantity

    Solution

    Work is a dot product of force and displacement and hence a scalar.

  • Question 2
    1 / -0

    SI unit of coefficient of viscosity is:

    Solution
    \[
    \begin{array}{l}
    \text { Coefficient of viscosity }=\frac{F r}{A v}=\frac{\left(M^{1} L^{1} T^{-2}\right)(L)}{\left(L^{2}\right)\left(L^{1} T^{-1}\right)}=M^{1} L^{-1} T^{-1} \\
    \frac{P a s c a l}{S}=\frac{M^{1} L^{-1} T^{-2}}{T^{1}}=M^{1} L^{-1} T^{-3} \\
    P a s c a l-s=M^{1} L^{-1} T^{-2} \times T^{1}=M^{1} L^{-1} T^{-1} \\
    N / m^{2} / \text { unit velocity }=\frac{M^{1} L^{1} T^{-2}}{L^{2} L T^{-1}}=M^{1} L^{-2} T^{-1} \\
    N / m / \text { unit velocity grad }=\frac{M^{1} L^{1} T^{-2}}{L\left(\frac{L T^{-1}}{L}\right)}=M^{1} L^{0} T^{-1}
    \end{array}
    \]
    Hence, option B is correct.
  • Question 3
    1 / -0

    Two masses m1 = 10kg and m2 = 5kg are connected by an ideal string as shown in the figure.

    The coefficient of friction between m1 and the surface is μ = 0.2. Assuming that the system is released from rest. The velocity of blocks when m2 has descended by 4m is (g = 10m/s2)

    Solution
    From constraint equations
    \[
    \begin{array}{l}
    v_{m_{1}}=v_{m_{2}}=v \\
    d_{m_{1}}=h_{m_{2}}=4 m
    \end{array}
    \]
    Using the work energy theorem,
    \(\Rightarrow E_{i}-E_{f}=\) Work done against friction
    \[
    \begin{array}{l}
    0-\left[\frac{1}{2} \times(10+5)\left(v^{2}\right)-5 \times 10 \times 4\right] \\
    =0.2 \times 10 \times 10 \times 4
    \end{array}
    \]
    Solving we get,
    \[
    v=4 m / s
    \]
  • Question 4
    1 / -0

    A projectile shot at an angle of 45o above the horizontal strikes the wall of a building 30m away at a point 15m above the point of projection. Initial velocity of the projectile is

    (take g=9.8m/s2)

    Solution
    Equation of trajectory:
    \[
    y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}
    \]
    The point (30,15) satisfies this equation.
    \[
    \begin{array}{l}
    15=30 \tan 45^{\circ}-\frac{9.8 \times 30^{2}}{2 u^{2} \cos ^{2} 45^{\circ}} \\
    \Rightarrow u^{2}=\frac{8820}{15}=588 \Rightarrow u=14 \sqrt{3}
    \end{array}
    \]
  • Question 5
    1 / -0

    The velocity of a particle depends upon as v = a + bt +ct2; if the velocity is in m/sec, the unit of a will be

    Solution

    Quantities of similar dimensions can be added or subtracted so unit of a will be same as that of velocity.

  • Question 6
    1 / -0

    A car starting from a point travels towards east with a velocity of 36 kmph. Another car starting from the same point travels towards north with a velocity of 24 kmph. The magnitude of relative velocity of one with respect to another is

    Solution
    relative velocity \(=36 \mathrm{i}-24 \mathrm{j}\) modulus of relative velocity \(=\sqrt{144(9+4)}=12 \sqrt{13} \mathrm{kmph}\) (because the relative velocities will be 24 and 36 in perpendicular directions)
  • Question 7
    1 / -0

    A small sphere of mass m is attached to a spring of spring factor k and normal length l. If the sphere rotates with radius r at frequency v then tension in the spring is :

    Solution
    Here the tension in the spring is the centripetal force, i.e \(T=m w^{2} r\)
    As \(\nu\) is the frequency so \(w=2 \pi \nu\) \(\therefore T=m r(2 \pi \nu)^{2}\)
  • Question 8
    1 / -0

    A body moving in a circular path with a constant speed has a ___

    Solution

  • Question 9
    1 / -0

    The following motion is based on the law of conservation of angular momentum

    A) rotation of top

    B) diving of a diver

    C) rotation of ballet dancer on smooth horizontal surface

    D) a solid sphere that rolls down on an inclined plane

    Identify the wrong statement:

    Solution

    In rotating a top, diving of a diver and rotation of a dancer, there is no external torque involved. Thus angular momentum is conserved.

    But when a sphere rolls down an incline, external torque (torque due to friction) is involved.

  • Question 10
    1 / -0
    A small satellite is revolving near earth's surface. Its orbital velocity will be nearly
    Solution
    Answer:
    These satellite have \(v_{0}=\sqrt{\frac{G M}{R}}=\sqrt{9 R}=8 \mathrm{km} / \mathrm{s}\)
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