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Physics Test 220

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Physics Test 220
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Weekly Quiz Competition
  • Question 1
    4 / -1

    In a series LCR circuit, R = 200Ω and the voltage and frequency of the main supply are 220 V and 50 Hz, respectively. On taking out the capacitance from the circuit, the current lags behind the voltage by 30°. On taking out the inductor from the circuit, the current leads the voltage by 30°. The power dissipated in the LCR circuit is

    Solution

     

  • Question 2
    4 / -1

    Two equally charged identical metal spheres A and B repel each other with a force of 2.0 ×10–5 N. Another identical uncharged sphere C is made to touch A and then placed at the midpoint between A and B. What is the net force on C?

    Solution

     

  • Question 3
    4 / -1

    What is the use of ball bearings between hubs and the axle of ceiling fans and bicycles?

    Solution

    Since the rolling friction is smaller than the sliding friction, thus, sliding friction is converted into rolling friction with the use of ball bearings.

     

  • Question 4
    4 / -1

    A particle of mass 0.3 kg is subjected to a force F = -kx, with k = 15 N/m. What will be its initial acceleration if it is released from a point 20 cm away from the origin?

    Solution

    Initial acceleration is overcome by retarding force.

    Or, m × (acceleration, a) = 3

    Or, a = 3/m = 3/0.3 = 10 ms-2

     

  • Question 5
    4 / -1

    A projectile is fired at an angle 60° with some velocity u. If the angle is changed infinitesimally, let the corresponding fractional changes in the range and the time of flight be x and y, respectively, then find the value of y.

    Solution

     

  • Question 6
    4 / -1

    The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is

    Solution

     

  • Question 7
    4 / -1

    A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0 cm. In normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective? (The diameter of the moon is 3.5 × 106 m and the radius of the lunar orbit around the Earth is 3.8 × 108 m)

    Solution

    Since u >> f0, v = f0 = 19 m.

    Now, u = -3.8 × 108 m

    Therefore, magnification produced by the objective,

    ∴ Diameter of the image of the moon = 3.5 × 106 0.5 × 10-7 = 0.175 m = 17.5 cm

    Hence, the correct choice is (4).

     

  • Question 8
    4 / -1

    Carbon, silicon and germanium have four valence electrons each. At room temperature, which of the following statements is the most appropriate?

    Solution

    C, Si and Ge have the same lattice structure and their valence electrons are 4. For C, these electrons are in the second orbit, for Si, they are in the third orbit, and for germanium, it is in the fourth orbit. In solid state, higher the orbit, greater the possibility of overlapping of energy bands. Ionisation energies are also less, therefore Ge has more conductivity as compared to Si. Both are semiconductors. Carbon is an insulator.

     

  • Question 9
    4 / -1

    A metallic container is completely filled with a liquid. The coefficient of linear expansion of the metal is 2.0 × 10-6 per °C and the coefficient of cubical expansion of the liquid is 6.0 × 10-6 per °C. On heating the vessel,

    Solution

    The coefficient of cubical expansion of the vessel = 3α = 3 × 2.0 × 10-6 = 6.0 × 10-6 per °C.

    Hence, 3α = γ

    Hence, cubical expansion of the vessel is equal to the cubical expansion of the liquid. Hence, the increase in the volume of the liquid is equal to the increase in the volume of the vessel. Thus, the correct choice is (3).

     

  • Question 10
    4 / -1

    Directions: In this question, two statements are given, numbered 1 and 2. Read both the statements carefully and choose the correct option accordingly.

    Statement-1: The temperature dependence of resistance is usually given as R = R0(1 + αΔt). The resistance of a wire changes from 100 Ω to 150 Ω when its temperature is increased from 27°C to 227°C. This implies that α = 2.7 10-3/°C.

    Statement-2: R = R0(1 + αΔt) is valid only when the change in the temperature ΔT is small and ΔR = (R - R0) << R0.

    Solution

    The resistance of a wire changes from 100Ω to 150Ω when the temperature is increased from 27°C to 227°C.

    It is true that α is small. But (150 - 100)Ω or 50Ω is not much less than 100Ω , i.e. R - R0 < r0="" is="" not="" />

     

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