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Physics Test - 15

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Physics Test - 15
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  • Question 1
    1 / -0

    A monochromatic light source of wavelength λ is placed at S. Three slits S1, S2 and S3 are equidistant from the source S and the point P on the screen. S1P-S2P=λ/6 and S1P-S3P = 2λ/3. If I be the intensity at P when only one slit is open, the intensity at P when all the three slits are open is

    Solution

  • Question 2
    1 / -0

    A luminous point object is moving along the principal axis of a concave mirror of focal length 12 cm towards it. When its distance from the mirror is 20 cm its velocity is 4 cm/s. The velocity of the image in cm/s at that instant.

    Solution

  • Question 3
    1 / -0

    An ammeter is designed by shunting a 60 Ω galvanometer with a 60 Ω resistance. The additional shunt that should be connected across it to double the range is

    Solution
    The shunt for ammeter
    \(S =\left\{\frac{ I _{ g }}{ I - I _{ g }}\right\} \times G\)
    \(\therefore \frac{ G }{ S }=\left(\frac{ I - I _{ g }}{ I _{ g }}\right)=\left(\frac{1}{ l _{ g }}-1\right)\)
    \(\frac{60}{60}=\frac{I}{I_{g}}-1\) or \(\frac{I}{I_{g}}=2\)
    \(\Rightarrow I=2 I_{g}\)
    When the range is doubled, \(1=4 l_{g}\) the shunt required, \(S =\left\{\frac{ I _{ g }}{4 I _{ g }- I _{ g }}\right\} \times 60=\frac{60}{3}=20 \Omega\)
    Now initial shunt resistance is \(60 .\) Another shunt let have resistance \(R\). It will be connected parallely with first resistance and equivalent resistance become 20. So, \(R_{e q}=\frac{60 R}{60+R}\)
    \(\Rightarrow 20=\frac{60 R}{60+R}\)
    \(\Rightarrow R=30 \Omega\)
    Hence option C is the correct answer.
  • Question 4
    1 / -0

    Two blocks P and Q are connected by a light inextensible string passing over a smooth pulley fixed as shown in the Figure. The coefficient of friction of blocks P and Q to the table is μ = 0.3; the mass of the block Q is 20 kg. The mass of the block P for the block just to slip is

    Solution

  • Question 5
    1 / -0

    One million small identical drops of water, all charged to the same potential, are combined to form a single large drop. If E is the sum of the electrostatic energy of each small drop, the combined energy of the large drop is

    Solution
    Energy of one drop \(=\frac{1}{2} C V^{2}=\frac{1}{2} q V\)
    where \(q= CV\) is the charge of one drop, \(C\) is the capacitance. Energy of \(10^{6}\) drops,
    \(E=\frac{1}{2}\) qV \(10^{6}\) Potential on the combined drop \(=(\) potential of each initial drop \() \times\left(n^{2 / 3}\right)\) \(V^{\prime}=V\left(10^{6}\right)^{2 / 3}=10^{4} V\)
    Energy of combined drop, \(E ^{\prime}=\frac{1}{2} q ^{\prime} V ^{\prime}=\frac{1}{2}\left(10^{6} q \right) V ^{\prime}\)
    \(=\frac{1}{2} 10^{6} q 10^{4} V\)
    \(=10^{4} E\)
    Hence option D is the correct answer.
  • Question 6
    1 / -0

    A square lead slab of side a = 50 cm and thickness t = 0.5 cm is subjected to a shearing force of magnitude F = 9 × 104 N. If the shear modulus of lead is η = 5.6 × 109 Pa, then how much work is done by the force?

    Solution
    \(\eta=\frac{\text { Shearing stress }}{\text { Shearing strain }}\)
    Work done by the force shear stress \(\times\) shear strain \(\times\) volume
    \(=\frac{1}{2} \times \frac{ F }{ a ^{2}} \times \frac{ F }{ a ^{2} \eta} \times a ^{2} t\)
    \(=\frac{F^{2} t}{2 a^{2} \eta}=\frac{\left(9 \times 10^{4}\right)^{2}\left(0.5 \times 10^{-2}\right)}{2(0.5)^{2} \times\left(5.6 \times 10^{9}\right)}\)
    \(=1.44 \times 10^{-2} J\)
    Hence option D is the correct answer.
  • Question 7
    1 / -0

    E, m, p and G denote energy, mass, angular momentum and gravitational constant. Then Ep2m5G2has the dimensions of

    Solution
    \(E=M L^{2} T^{-2} ; m=M ; p=M L^{2} T^{-1}\)
    \(G=M^{-1} L^{3} T^{-2}\)
    \(\therefore \frac{E p^{2}}{m^{5} G^{2}}=\frac{M L^{2} T^{-2} \cdot M^{2} L^{4} T^{-2}}{M^{5} \cdot M^{-2} L^{6} T^{-4}}\) = no dimension
    \(\therefore\) it represents angle which has no dimension.
    Hence option C is the correct answer.
  • Question 8
    1 / -0

    Consider telecommunication through optical fibres. Which of the following statements is NOT true?

    Solution

    Some of the characteristics of an optical fibre are as follows

    (i) This works on the principle of total internal reflection.

    (ii) It consists of core made up of glass/silica/plastic with refractive index n1, which is surrounded by a glass or plastic cladding with refractive index n2 (n2 > n1). The refractive index of cladding can be either changing abruptly or gradually changing (graded index fibre).

    (iii) There is a very little transmission loss through optical fibres.

    (iv) There is no interference from stray electric and magnetic field to the signals through optical fibres.

    Hence option B is the correct answer.

  • Question 9
    1 / -0

    The number of AM broadcast stations that can be accommodated at a 150 kHz band width, if the highest frequency modulating carrier is 5 kHz is

    Solution

    Total bandwidth = 150 kHz

    fa(max) = 5 kHz

    Any station being modulated by a 5 kHz will produce an upper-side frequency 5 kHz above its carrier and a lower-side frequency 5 kHz below its carrier.

    Thus one station needs a bandwidth of 10 kHz.

    Number of stations accommodated = Total BW /BW per station

    =(150×103)/(10×103)=15

    Hence option A is the correct answer.

  • Question 10
    1 / -0

    The frequency of turning forks X and Y are respectively 15% more and 8% less than the frequency of turning fork Z. When X and Y are simultaneously excited, 45 beats per second are produced. Find the frequency of turning fork X is

    Solution
    Let the frequency of fork \(Z\) is \(f_{z}\). Then \(f_{X}=f_{z}+\frac{15 f_{z}}{100}=\frac{115 f_{z}}{100}\)
    \(f_{T}=f_{z}-\frac{8 f_{z}}{100}=\frac{92 f_{z}}{100}\)
    \(f_{x}-f_{y}=45\)
    \(\frac{23 f_{2}}{100}=45\)
    \(f_{2}=\frac{45 \times 100}{23}=195.6\)
    So frequency of fork \(X\) is \(f_{X}=\frac{115}{100} \times 195.6=225\)
    Hence option C is the correct answer.
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