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Physics Test - 15

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Physics Test - 15
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  • Question 1
    4 / -1
    The galvanometer deflection, when key \({K}_{1}\) is closed but \({K}_{2}\) is open, equals \(\theta_{0}\) (see figure). On closing \({K}_{2}\) also and adjusting \({R}_{2}\) to \(5 \Omega,\) the deflection in galvanometer becomes \(\frac{\theta_{0}}{5} .\) The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]:
    Solution

    Case 1 : In the first case, it is given that, when the key \(K_{1}\) is closed but \(K_{2}\) is open then deflection in the galvanometer equals \(\theta_{0}\).

    Let the current flowing in the galvanometer is \(i_{g}\) and the deflection in the galvanometer is given by \(\theta_{0}\). We know that the current in the galvanometer is proportional to the deflection in the galvanometer.

    Thus, \(i_{g} \propto \theta_{0}\)

    After removing proportional sign we get a constant C,

    \(i_{g}=C \theta_{0}\)......(1)

    Let us assume the emf of the battery is \(E\).

    We know that the current flowing in the circuit is equal to the ratio of emf of the battery to the total resistance in the circuit.

    So, \(i_{g}=\frac{E}{220+R_{g}}\)......(2)

    From equation (1) and (2), we get

    \(\frac{E}{220+R_{g}}=C \theta_{0}\)......(3)

    Now, according to the second condition when \(K_{2}\) is closed and adjusting \(R_{2}\) to \(5 \Omega\), the deflection in the galvanometer becomes \(\frac{\theta_{0}}{5}\).

    We know that the current in the galvanometer is proportional to the deflection in the galvanometer. Here deflection is\(\frac{\theta_{0}}{5} \)

    \(i_{g}=C \frac{\theta_{0}}{5}\)........(4)

    We know that the current flowing in the circuit is equal to the ratio of emf of the battery to the total resistance in the circuit.

    \(i_{g}=\left(\frac{E}{220+\frac{5 R_{g}}{5+R_{g}}}\right) \times\left(\frac{5}{R_{g}+5}\right)\)...........(5)

    From equation (4) and (5), we get

    \(\left(\frac{E}{220+\frac{5 R_{g}}{5+R_{g}}}\right) \times\left(\frac{5}{R_{g}+5}\right)=C \frac{\theta_{0}}{5}\)

    On further solving this, we get

    \(\Rightarrow \frac{5 E}{225 R_{g}+1100}=\frac{C \theta_{0}}{5}\).......(6)

    Now finally solving equation (3) and (6) from case 1 and 2 respectively, we get,

    \(\Rightarrow \frac{225 R_{g}+1100}{1100+5 R_{g}}=5\)

    On further solving, we get

    \(\Rightarrow 5500+25 R_{g}=225 R_{g}+1100\)

    \(\Rightarrow 200 R_{g}=4400\)

    On finally solving this, we get

    \(R_{g}=22 \Omega\)

    Thus, the resistance of the galvanometer is \(22 \Omega\).

  • Question 2
    4 / -1

    Due to an acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\), the velocity of a body increases from \(20\) \(\mathrm{m} / \mathrm{s}\) to \(30 \mathrm{~m} / \mathrm{s}\) in a certain period. Find the displacement (in \(\mathrm{m}\) ) of the body in that period.

    Solution

    Given -

    Acceleration \((\mathrm{a})=2 \mathrm{~m} / \mathrm{s}^{2}\)

    Final velocity \((\mathrm{v})=30 \mathrm{~m} / \mathrm{s}\)

    Initial velocity \((\mathrm{u})=\) \(20 \mathrm{~m} / \mathrm{s}\)

    Displacement \(=\mathrm{s}\)

    We know,

    \(\Rightarrow v^{2}-u^{2}=2 a s\) (Equation of Motion)

    \(\Rightarrow s=\frac{(v^{2}-u^{2})}{(2 \times a)}\)

    \(\Rightarrow s=\frac{(30^{2}-20^{2})}{(2 \times 2)}\)

    \(\Rightarrow s=\frac{500}{4}\)

    \(\Rightarrow s=125 m\)

  • Question 3
    4 / -1

    The empty space between the plates of a capacitor is filled by a liquid of dielectric constant K, the capacitance of capacitor:

    Solution

    When the empty space between the plates of a capacitor is filled by a liquid of dielectric constant K, the capacitance of the capacitor increased by the factor of K.

    The capacitance C = K C0 where C0 is the capacitance without the dielectric.

  • Question 4
    4 / -1

    A wooden box of mass \(0.2\) kg is suspended from the ceiling by a light string. A second wooden box of mass \(0.3\) kg is suspended from the first box by another string. Find the tensions in both the strings. Take \(g=10\) m/s.

    Solution

    The free-body diagrams for both the boxes are given below:

  • Question 5
    4 / -1
    A modulated signal \(\mathrm{C}_{\mathrm{m}}(\mathrm{t})\) has the form \(\mathrm{C}_{\mathrm{m}}(\mathrm{t})=30 \sin 300 \pi \mathrm{t}+10(\cos 200 \pi \mathrm{t}-\cos 400 \pi \mathrm{t})\)The carrier frequency \(c\), the modulating frequency (message frequency) \(\mathrm{f}_{\omega}\), and the modulation index \(\mu\) are respectively given by:
    Solution
    \(C_{m}(t)=30 \sin 300 \pi t+10(\cos 200 \pi t-\cos 400 \pi t)\)
    Standard equation of amplitude modulated wave,
    \(\mathrm{C}_{\mathrm{m}}(\mathrm{t})=\mathrm{A}_{\mathrm{c}} \sin \left(\omega_{\mathrm{c}} \mathrm{t}\right)-\frac{\mu A_{c}}{2} \cos \left(\omega_{c}+\omega_{m}\right) \mathrm{t}+\frac{\mu A_{c}}{2} \cos \left(\omega_{c}-\omega_{m}\right) \mathrm{t}\)
    By comparing we get,
    \(A_{c}=30 \mathrm{~V}\)
    \(\omega_{\mathrm{C}}=300 \pi\)
    \(\Rightarrow 2 \pi f_{c}=300 \pi\)
    \(\Rightarrow \mathrm{f}_{\mathrm{c}}=150 \mathrm{~Hz}\)
    \(\omega_{\mathrm{C}}-\omega_{\mathrm{S}}=200 \pi\)
    \(\Rightarrow 2 \pi\left(\mathrm{f}_{\mathrm{c}}-\mathrm{f}_{\omega}\right)=200 \pi\)
    \(\Rightarrow \mathrm{f}_{\mathrm{c}}-\mathrm{f}_{\omega}=100 \mathrm{~Hz}\)
    \(\therefore f_{\omega}=150-100=50 \mathrm{~Hz}\)
    and \(\frac{\mu A_{c}}{2}=10\)
    \(\Rightarrow \mu \times \frac{30}{2}=10\)
    \(\Rightarrow \mu=\frac{10}{15}=\frac{2}{3}\)
  • Question 6
    4 / -1

    When a beam of light passes through a prism, which of the following color(s) deviates more?

    Solution

    As we know, the bending of light is inversely proportional to the wavelength of the light.

    Bending \(\propto \frac{1}{\text { Wavelength }}\)

    Different colors bend differently on passing through a prism and the bending of colour depends on wavelength.

    Higher wavelengths bend less, whereas shorter wavelength bend more.

    • Violet: 380–450 nm.
    • Green: 495–570 nm.
    • Yellow: 570–590 nm.
    • Red: 620–750 nm.

    Out of all violet has the least wavelength and therefore, it bends the most and red has the highest wavelength, so it bends the least.

  • Question 7
    4 / -1

    Where does a body has the maximum weight?

    Solution

    The weight of an object is maximum at poles as the gravity is slightly higher at the poles and hence the weight there is more as compared to anywhere else.

  • Question 8
    4 / -1

    The motion of a body in \(x-y\) plane is represented by \(x=4-9 t\) and \(y=t^{2}\) where \(x, y\) are in metre. Find the magnitude of its absolute velocity on \(t=6\) sec.

    Solution

    Absolute velocity is given by:

    \(V=\sqrt{v_{x}^{2}+v_{y}^{2}}\)

    \(\mathrm{v}_{\mathrm{x}}=\) velocity component in \(x\) direction \(=\frac{d x}{d t}\)

    \(v_{y}=\) velocity component in \(y\) direction \(=\frac{d y}{d t}\)

    On \(t=6 \)sec

    \(x=4-9 t\)

    \(v_{x}=\frac{d x}{d t}=-9\)

    \(y=t^{2}\)

    \(v_{y}=\frac{d y}{d t}=2 t=2(6)=12\)

    \(V=\sqrt{(-9)^{2}+(12)^{2}}=\sqrt{225}=15\) m/s

  • Question 9
    4 / -1

    Two particles \(A\) and \(B\) have a phase difference of \(\pi\) when a sine wave passes through the region.

    (I) \(A\) oscillates at half the frequency of \(B\)

    (II) \(A\) and \(B\) move in opposite directions.

    (III) \(A\) and \(B\) must be separated by half of the wavelength.

    (IV) The displacements at \(A\) and \(A\) have equal magnitudes

    Select the correct option:

    Solution

    \(A\) and \(B\) have a phase difference of \(\pi\). So, when a sine wave passes through the region, they move in opposite directions and have equal displacement. They may be separated by an odd multiple of their wavelength.

    \(yA \rightarrow=Asin \omega t\)

    \(yB \rightarrow=B \sin \omega t+\pi\)

  • Question 10
    4 / -1

    One oscillation of a swinging pendulum occurs when the bob moves from X to Y and back to X again.

    Which of the following method would be the most accurate way to measure the time for one oscillation of the pendulum using a stopwatch?

    Solution

    It is not an accurate method to time just one oscillation. The human reaction time of 0.3 s causes a large error in the reading. The best way is to time 20 oscillation and divides the time by 20 .

    If it takes 5 s for the pendulum to go back and forth 20 times. The time period of the pendulum is 0.25 s.

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