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Basic Science Test 12

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Basic Science Test 12
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  • Question 1
    1 / -0
    A body is projected with velocity 'u' so that the maximum height is thrice the horizontal range. Then the maximum height is
    Solution
    $$H_{max} = \dfrac{4^2sin^2\theta}{2g}$$
    Range $$= \dfrac{4^2sin2\theta}{g} = \dfrac{4^2sin\theta\,cos\theta}{g}$$
    $$\dfrac{H}{3} = R = \dfrac{4sin2\theta}{2 \times 3g} = \dfrac{4sin\theta\,cos\theta}{g}$$
    $$tan\theta = \dfrac{12}{1}$$
    $$sin\theta = \dfrac{12}{\sqrt{12^2 + 1^2}} = \dfrac{1}{\sqrt{145}}$$
    $$H_{max} = \dfrac{4^2sin^2\theta}{2g} = \dfrac{4^2\left ( \dfrac{12}{\sqrt{145}} \right )^2}{2g}$$
    $$= \dfrac{4^2 \times 144}{145 \times 2 \times g} = \dfrac{72u^2}{145\,g}$$
  • Question 2
    1 / -0
    A body projected with velocity $$30\,ms^{-1}$$ reaches its maximum height in $$1.5\,s.$$ Its range is
    Solution
    $$T = \dfrac{2u \,sin\theta}{g}$$
    $$\dfrac{t}{2} = 1.5 = \dfrac{u\,sin\theta}{g}$$
    $$sin\theta = \dfrac{1.5 \times g}{u}$$
    $$= \dfrac{1.5 \times 10}{30}$$
    $$sin\theta = \dfrac{1}{2} \Rightarrow \theta = 30^\circ$$
    $$R = \dfrac{u^2sin2\theta}{g}$$
    $$= \dfrac{(30)^2sin60^\circ}{10}$$
    $$= 45\sqrt{3}\,m$$
  • Question 3
    1 / -0
    A body of 10 Kg is suspended by a rope and is pulled to a side by means of a horizontal force so that th rope makes an angle of $$60^{\circ}$$ with the vertical. The tension in the rope is 
    Solution
    $$T\sin 30 = mg = 10 \times 10$$
    $$t \times \dfrac{1}{2} = 100$$
    $$T = 200 N$$ or $$20 kg$$

  • Question 4
    1 / -0
    An objective move along $$x-$$axis such that its position varying with time $$t$$ is given as $$x=4t-t^2 $$ ($$4x$$ is in metre and time $$t$$ in second). The distance travelled by the object from $$t=0$$ to $$t=3\ s$$ is
    Solution

    Given,
    $$x=t^2 -4t+6$$
    So, the change in velocity is:
    $$\dfrac {dx}{dt}=2t-4$$
    Since velocity is changing,
    At $$t=0, x_1 =6$$
    At $$t=2, x_2 =2$$
    So the magnitude of the distance traveled is
    $$6-2=4m$$
    At $$t=3, x_3=3$$
    So distance traveled from $$t=240, t=3s$$
    $$x_3 -x_2=3-2=1m$$
    Thus the total distance is:
    $$4+1=5m$$
    $$x=4t-t^2$$
    $$V\Rightarrow 4-2t$$
    $$t=2$$
    $$\displaystyle \int_0^2 (4-2t)dt \displaystyle \int_2^3 (4-2t)dt$$
    $$(4t-t^2)^2_6 +(4t-t^2)_2^3$$
    $$4+1$$
    $$\to 5$$
  • Question 5
    1 / -0
    A wheel whose moment of inertia is $$12\ kg\ m^{2}$$ has an initial angular velocity of $$40\ rad/s$$. A constant torque of $$20\ Nm$$ acts on the wheel. The time in which the wheel is accelerated to $$100\ rad/s$$ is
  • Question 6
    1 / -0
    For a particle moving along a straight line, its position $$x$$ is given by $$x = 2e^{-2t}$$, where $$t$$ is time. Relation between its velocity $$'v'$$ and acceleration $$'a'$$ is given by
    Solution
    $$x = 2e^{2t}$$
    $$\dfrac {dx}{dt} = V\Rightarrow V = -4e^{-2t}$$
    $$\dfrac {d^{2}x}{dt^{2}} = \dfrac {dV}{dt} = a \Rightarrow a = 8e^{-2t}$$.
    $$a = 8e^{-2t} = -2(-4e^{-2t})$$
    $$= -2V$$ as $$V = (-4e^{-2t})$$
    $$\Rightarrow a = -2V$$.
  • Question 7
    1 / -0
    Two charges of $$1.0\ C$$ each are placed one meter apart in free space. The force between them will be
    Solution

  • Question 8
    1 / -0
    Two men of masses $$m$$ and $$m/2$$ starts climbing up on two massless strings fixed at the ceiling with acceleration $$g$$ and $$g/2$$ respectively. The ratio of tensions in the two straight will be
    Solution
    $$T_{1} = mg + mg$$
    $$T_{1} = 2mg$$
    $$T_{2} = \dfrac {m}{2}g + \dfrac {m}{2}\cdot \dfrac {g}{2}$$
    $$= \dfrac {m}{2}\left [\dfrac {3g}{2}\right ] = \dfrac {3mg}{4}$$
    $$\therefore \dfrac {T_{1}}{T_{2}} = \dfrac {2mg}{3mg} \times \dfrac {4}{1}$$
    $$= \dfrac {8}{3}$$.

  • Question 9
    1 / -0
    Two bodies are thrown with the same initial velocity of $$30\ m/s$$. One at $$17^{\circ}$$, other at $$73^{\circ}$$ to the horizontal. The sum of the maximum heights reached by them is
    Solution
    $$H_{1}=\dfrac{V^{2}\sin^{2}\theta}{2g}=\dfrac{(30)^{2}\times \sin^{2}17^{o}}{2\times 10}=122\sin^{2}7^{2}$$

    $$H_{2}=\dfrac{V^{2}\sin^{2}73^{o}}{2g}=45\sin^{2}73^{o}$$

    $$H_{1}+H_{2}=45(\sin^{2}17^{o}+\sin^{2}73^{o})$$

    $$=45[(\cos 73^{o})^{2}+\sin^{2}73^{o}]$$

    $$=45(\sin^{2}73^{o}+\cos^{2}73^{o})=45$$
  • Question 10
    1 / -0
    What is the angle between $$\vec{i} +\vec{j}$$ and $$\vec{i}$$?
    Solution
    We know that
    $$\cos \theta=\dfrac{|\vec{i} +\vec{j}.\vec{i}|}{|\vec{i} +\vec{j}||\vec{i}|}$$

    $$\cos \theta=\dfrac{1}{\sqrt{2}.1}=\dfrac{1}{\sqrt{2}}$$

    $$\theta=45^\circ$$
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