What is the effect on the image formed by the convex lens if the central portion of the lens is wrapped in black paper as shown in the figure?

• Correct Answer: C

  Solution :

   In the convex lens the rays are converted to a point on the other side lens when the parallel rays are incident on the surface of the lens. If the central portion of the lens wrapped by black paper, full image will be form but with a less brightness.

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• Correct Answer: D

  Solution :

  Given: Angle of the prism \[=30{}^\circ \] Refractive index \[\mu =\sqrt{2}\] The relation for refractive index is given by \[\mu =\frac{\frac{A+{{\delta }_{m}}}{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{{{30}^{o}}+{{\delta }_{m}}}{2}}{\sin \frac{\sin {{30}^{o}}}{2}}\] \[\sqrt{2}=\frac{\sin \frac{{{30}^{o}}+{{\delta }_{m}}}{2}}{\sin 150}\] or \[\sin =\frac{{{30}^{o}}+{{\delta }_{m}}}{2}=\sqrt{2}\sin {{15}^{o}}\] or \[\sin =\frac{{{30}^{o}}+{{\delta }_{m}}}{2}=1.414\times 0.2588\] or \[\sin =\frac{{{30}^{o}}+{{\delta }_{m}}}{2}=0.3659\times =0.366\]              or \[\frac{{{30}^{o}}+{{\delta }_{m}}}{2}={{\sin }^{-1}}0.366={{21.46}^{o}}={{21.5}^{o}}\] or \[{{30}^{o}}+{{\delta }_{m}}={{43}^{o}}\] \[{{\delta }_{m}}={{43}^{o}}-{{30}^{o}}={{13}^{o}}\]

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Page 3

• Correct Answer: C

  Solution :

   The molecules of a gas are in random motion always. According to the kinetic energy of gases, translational kinetic energy of the gas molecules for one mole of a gas is\[=\frac{3}{2}RT\]

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Page 4

• Correct Answer: A

  Solution :

   The radius of earth is reduced by =2% (given) the weight of body or gravitational force\[g=\frac{GM}{R_{e}^{2}}\propto \frac{1}{R_{e}^{2}}\] (where \[{{\operatorname{R}}_{e}}\] is the radius of the earth) Hence, if the radius of the earth is reduced then the value of g will be increased so that the weight of body will increase.

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Page 5

• Correct Answer: C

  Solution :

   The work done during the process from P to Q= area PQRSTP (positive sign is to be taken due to expansion along PQ) Area of triangle PQR + area of rectangle PRST \[=\frac{1}{2}\times 2\times 1\times 1\times 2=3PV\] Work done during process from R to P = - Area RSTP (negative sign due to compression in atom RP)\[=-ST\times PT=-2\times 1=-2PV\] Hence, the work done in the complete cycle \[=3PV-2PV=PV\]

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Page 6

• Correct Answer: B

  Solution :

   Given : Focal   length   of   objective   lens \[{{f}_{e}}=180\,cm\] Focal length of eye piece \[{{f}_{e}}=6\,cm\] In the  normal  adjustment,  the magnifying power of telescope is given by \[=\frac{{{f}_{0}}}{{{f}_{e}}}=\frac{180}{6}=30\]

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Page 7

• Correct Answer: A

  Solution :

   Distance between two consecutive nodes with air \[{{l}_{A}}=5cm.\] Distance between two consecutive nodes with hydrogen \[{{l}_{A}}=13.4cm.\] Velocity of sound in air = 330 m/s The velocity of sound is given by \[\upsilon H={{\upsilon }_{A}}\frac{{{l}_{H}}}{{{l}_{A}}}=330\times \frac{13.4}{5}\] \[=884.5m/s\]

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Page 8

• Correct Answer: A

  Solution :

   If the second harmonic or first overtone has to be generated in a string of length I stretched between to rigid supports. Then the points where the string has to be plucked is\[\frac{l}{4}\] and touched at \[\frac{l}{2}\]

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Page 9

• Correct Answer: A

  Solution :

   The given figure represented balanced Wheat, stone bridge since, \[\frac{A}{B}=\frac{2\Omega }{3\Omega }=1.5\Omega ,\frac{C}{D}=\frac{6\Omega }{4\Omega }=1.5\Omega \] Hence, the potential between R and S same therefore resistance in the arm RS of \[7\Omega \] resistance is in effective. The resistance in the upper arm are connected in series combination. Hence the equivalent resistance of the upper arm is given by \[{{R}_{V}}=2+3=5\Omega \] Similarly the resistance in the lower arm are also connected in the series combination, hence their equivalent resistance is \[{{R}_{L}}=4+6=10\Omega \] Now\[{{R}_{U}}\]and\[{{R}_{L}}\]are in parallel combination. Therefore the equivalent resistance is \[{{\operatorname{R}}_{eq}}=\frac{5\times 10}{5+10}=\frac{50}{15}=\frac{10}{3}\Omega \]

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Page 10

• Correct Answer: D

  Solution :

   Given : Initial capacitance = C Initial thickness of the sheet \[{{t}_{1}}=d\](where d is the separation between plates) Final thickness of sheet \[{{t}_{2}}=\frac{d}{2}\] The capacitance of a parallel plate capacitor is given by \[C=\frac{{{E}_{0}}A}{d}\propto \frac{1}{d}\] Hence \[\frac{{{C}_{1}}A}{{{C}_{2}}}=\frac{{{d}_{2}}}{{{d}_{1}}}=\frac{2}{d}=\frac{1}{2}\](where \[{{C}_{2}}\]is the new capacitance of a capacitor) Hence, \[{{C}_{2}}=2{{C}_{1}}=2C\]

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Page 11

• Correct Answer: B

  Solution :

   When the forward voltage in a diode is increased, the effective barrier voltage is reduced. If increased the current across the junction. On this account the width of depletion layer is decreased.

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Page 12

• Correct Answer: B

  Solution :

   As we are quite well know that the power of an electric bulbs is \[P=Vi=V\frac{V}{R}=\frac{{{V}^{2}}}{R}\] Hence  \[\frac{P}{{{P}_{0}}}=\frac{{{V}^{2}}}{R}\times \frac{R}{V_{0}^{2}}={{\left( \frac{V}{{{V}_{0}}} \right)}^{2}}\] Thus   \[P={{P}_{0}}{{\left( \frac{V}{{{V}_{0}}} \right)}^{2}}\]

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Page 13

• Correct Answer: C

  Solution :

   When the magnet falls into the coil of a conducting wire the magnetic flux linked with the ring charges, and induced current is set up in the ring which opposes the downward motion of the magnet. Hence the acceleration of the magnet will be less than the acceleration due to gravity.

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Page 14

• Correct Answer: A

  Solution :

   First resistance \[{{R}_{1}}=2\Omega \] Second resistance \[{{R}_{2}}=4\Omega \] Third resistance \[{{R}_{3}}=8\Omega \] (given) These three resistances are connected in parallel. As the relation for the equivalent resistance of parallel combination is given by \[\frac{1}{R}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\] \[=\frac{7}{8}\] Thus \[R=\frac{8}{7}\Omega \]

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Page 15

• Correct Answer: A

  Solution :

   Given : Potential at A, V= 600 volt mass of the ball \[m=1\,gm={{10}^{-3}}kg\]charge on the ball \[q={{10}^{-8}}C\]velocity at\[B,{{\upsilon }_{B}}=20cm/s\] Now from the relation \[\upsilon _{A}^{2}-\upsilon _{B}^{2}=\frac{2qV}{m}=\frac{2\times {{10}^{-8}}\times 600}{{{10}^{-3}}}=0.012\] Thus \[\upsilon _{A}^{2}=0.012+0.04=0.052\] \[{{\upsilon }_{A}}=0.22.8cm/\sec .\] \[{{\upsilon }_{A}}=22.8\,cm/\sec .\]

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Page 16

• Correct Answer: C

  Solution :

   From the figure, the distance between the observer and the image is\[=30+10=40cm\]
  

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• Correct Answer: A

  Solution :

   Given : Range of milliammeter \[{{A}_{ig}}=10mA=10\times {{10}^{-3}}amp.\] Resistance \[G=1\Omega \] Range of voltmeter V = 10 volt The relation for the resistance that must be connected is given by \[=\frac{V}{ig}-G=\left( \frac{10}{10\times {{10}^{-3}}} \right)-1=1000-1=999\Omega \]

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Page 18

• Correct Answer: B

  Solution :

   While a capacitor remains connected to a battery and dielectric slab is slipped between  the  plates,  the potential difference between the plates remain uncharged. The introduction of dielectric slab increases the charge of capacitor which flows from the battery.

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Page 19

• Correct Answer: C

  Solution :

    Given : Power of first bulb \[{{P}_{1}}=40\]watt Power of second bulb \[{{P}_{2}}=40\] watt As the power of the bulb is given by \[P=\frac{V}{R}\propto \frac{1}{R}\] Hence \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}}.\]or\[\frac{40}{60}=\frac{{{R}_{2}}}{{{R}_{1}}}\]or\[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{60}{40}=\frac{3}{2}\]or\[{{R}_{1}}:{{R}_{2}}=3:2\]

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Page 20

• Correct Answer: A

  Solution :

   The force on Q at A due to Q at B is given by \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{QQ}{{{a}^{2}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}^{2}}}{{{a}^{2}}}\] ?(1)
  
  the force on Q at C due to q is given by \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{{{\left( \frac{a}{2} \right)}^{2}}}\] ...(2) These forces should be equal and opposite for the equilibrium of charge Q at A. This. is only posible when q is negative \[\frac{{{Q}^{2}}}{{{a}^{2}}}=-\frac{Qq}{{{(a/2)}^{2}}}\]or\[q=-\frac{Q}{4}\] Thus\[q=-\frac{Q}{4}\]

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Page 21

• Correct Answer: B

  Solution :

   When a ball is thrown vertically upwards its velocity goes on decreasing due to gravitational acceleration till it becomes zero. After that it falls down with the same gravitation acceleration. Its velocity the point of projection becomes equal to the velocity of projection. Hence the time of ascent is equal to the time of descent.

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Page 22

• Correct Answer: C

  Solution :

   The work function \[{{W}_{0}}=4.125eV=4.125\times 1.6\times {{10}^{-19}}\]joule The relation for work function is\[{{W}_{0}}=\frac{hc}{\lambda }\](where \[\lambda \] is the cut off wavelength) \[6.6\times {{10}^{-19}}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{\lambda }\] \[\lambda =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{6.6\times {{10}^{-19}}}\] \[=\frac{19.8\times {{10}^{-26}}}{6.6\times {{10}^{-19}}}=3\times {{10}^{-7}}cm.\] \[=300\times {{10}^{-9}}m=300nm\]

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Page 23

• Correct Answer: B

  Solution :

   From the formula \[n=\frac{\upsilon +{{\upsilon }_{0}}}{\upsilon -\upsilon }n\] \[=\frac{335+5}{335-5}\times 165=170Hz\]
  
  Hence no. of beats heard per second by the passenger is \[=170-165=5\]beats/sec

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Page 24

• Correct Answer: A

  Solution :

   Critical angle \[C=30{}^\circ \] (given) From the law of total internal reflection we have \[\sin C=\frac{\upsilon }{c}\] \[\sin {{30}^{o}}=\frac{\upsilon }{c}\] \[0.5=\frac{\upsilon }{3\times {{10}^{8}}}\] \[\upsilon =0.5\times 3\times {{10}^{8}}=1.5\times {{10}^{8}}m/s\]

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Page 25

• Correct Answer: B

  Solution :

   As in the stationary wave, all the particle of the medium cross the mean position with different velocities at same instant.

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Page 26

• Correct Answer: C

  Solution :

   Volume of first cylinde \[{{V}_{1}}=5\] litres Volume of second cylinder \[{{V}_{2}}=30\] litre Hence, total volume \[{{V}_{2}}={{V}_{1}}+{{V}_{2}}=5+30=35\] litre The initial pressure \[{{p}_{1}}\] at NTP \[{{p}_{1}}=76cm\] of Hg Now from the Boyles law, \[{{p}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\] Hence \[{{p}_{2}}=\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right){{p}_{1}}\] \[=\frac{5}{35}\times 76=10.85cm\]of Hg

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