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Two bodies of masses M1 and M2 are connected to a ceiling as illustrated in the diagram below. The tension T2 in the middle string is given by the expression [Take acceleration due to gravity as g]

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14. Two bodies of masses M1M_1 and M2M_2 are connected to a ceiling as illustrated in the diagram below. The tension T2T_2 in the middle string is given by the expression [Take acceleration due to gravity as gg]

  1. Analyze the forces acting on mass M2M_2.

    The forces acting on M2M_2 are its weight acting downwards and the tension T2T_2 acting upwards.

  2. Apply Newton's second law.

    Since the system is in equilibrium (not accelerating), the net force on M2M_2 is zero.

    T2−M2g=0T_2 - M_2g = 0 T2=M2gT_2 = M_2g
    • The correct option is B.

    15. Two identical cells each of emf 2 V2 \text{ V} and internal resistance 0.5Ω0.5 \Omega are connected in series with a 4.0Ω4.0 \Omega resistor. Calculate the current in the circuit.

  3. Calculate the total electromotive force (emf) of the cells in series.

    Etotal=E1+E2=2 V+2 V=4 VE_{total} = E_1 + E_2 = 2 \text{ V} + 2 \text{ V} = 4 \text{ V}

  4. Calculate the total internal resistance of the cells in series.

    rtotal=r1+r2=0.5Ω+0.5Ω=1.0Ωr_{total} = r_1 + r_2 = 0.5 \Omega + 0.5 \Omega = 1.0 \Omega

  5. Calculate the total resistance in the circuit.

    Rcircuit=Rexternal+rtotal=4.0Ω+1.0Ω=5.0ΩR_{circuit} = R_{external} + r_{total} = 4.0 \Omega + 1.0 \Omega = 5.0 \Omega

  6. Use Ohm's Law to calculate the current.

    I=EtotalRcircuitI = \frac{E_{total}}{R_{circuit}} I=4 V5.0ΩI = \frac{4 \text{ V}}{5.0 \Omega} I=0.8 AI = 0.8 \text{ A}
    • The correct option is C.

    16. The first ionization energy of Titanium (Ti) is 6.36×107 J mol−16.36 \times 10^7 \text{ J mol}^{-1}. Calculate the longest wavelength of electromagnetic radiation that can ionize an atom of Ti. [Planck's constant, h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s}, speed of light, c=3.0×108 m s−1c = 3.0 \times 10^8 \text{ m s}^{-1}, Avogadro's number, NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23} \text{ mol}^{-1}]

  7. Convert the ionization energy from per mole to per atom.

    Eatom=6.36×107 J mol−16.02×1023 mol−1E_{atom} = \frac{6.36 \times 10^7 \text{ J mol}^{-1}}{6.02 \times 10^{23} \text{ mol}^{-1}} Eatom≈1.05648×10−16 JE_{atom} \approx 1.05648 \times 10^{-16} \text{ J}

  8. Use the formula for the energy of a photon, E=hcλE = \frac{hc}{\lambda}, to find the wavelength λ\lambda.

    λ=hcEatom\lambda = \frac{hc}{E_{atom}}

  9. Substitute the given values into the formula.

    λ=(6.63×10−34 J s)×(3.0×108 m s−1)1.05648×10−16 J\lambda = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m s}^{-1})}{1.05648 \times 10^{-16} \text{ J}} λ=1.989×10−25 J m1.05648×10−16 J\lambda = \frac{1.989 \times 10^{-25} \text{ J m}}{1.05648 \times 10^{-16} \text{ J}} λ≈1.8826×10−9 m\lambda \approx 1.8826 \times 10^{-9} \text{ m}

  10. Convert the wavelength to nanometers (nm).

    1 nm=10−9 m1 \text{ nm} = 10^{-9} \text{ m} λ≈1.8826 nm\lambda \approx 1.8826 \text{ nm}

    Rounding to two decimal places, λ≈1.88 nm\lambda \approx 1.88 \text{ nm}.

    • The correct option is B.

    A body accelerates uniformly from rest at 2 m s−22 \text{ m s}^{-2} for 10 s10 \text{ s}. Use the information provided above to answer questions 17 and 18.
    Given:
    Initial velocity, u=0 m s−1u = 0 \text{ m s}^{-1} (from rest)
    Acceleration, a=2 m s−2a = 2 \text{ m s}^{-2}
    Time, t=10 st = 10 \text{ s}

    17. The speed of the body at the end of 10 s10 \text{ s} is

  11. Use the kinematic equation v=u+atv = u + at.

    v=0 m s−1+(2 m s−2)×(10 s)v = 0 \text{ m s}^{-1} + (2 \text{ m s}^{-2}) \times (10 \text{ s}) v=20 m s−1v = 20 \text{ m s}^{-1}
    • The correct option is C.

    18. How far will the body move in 10 s10 \text{ s}?

  12. Use the kinematic equation s=ut+12at2s = ut + \frac{1}{2}at^2.

    s=(0 m s−1)×(10 s)+12×(2 m s−2)×(10 s)2s = (0 \text{ m s}^{-1}) \times (10 \text{ s}) + \frac{1}{2} \times (2 \text{ m s}^{-2}) \times (10 \text{ s})^2 s=0+1×(100 s2) m s−2s = 0 + 1 \times (100 \text{ s}^2) \text{ m s}^{-2} s=100 ms = 100 \text{ m}
    • The correct option is C.

    19. The magnetized needle of a compass aligns more quickly at the geographical north and south poles than other regions because the poles have the
    The Earth's magnetic field lines are most concentrated at the magnetic poles, meaning the magnetic field strength is greatest there. This stronger field exerts a greater torque on the compass needle, causing it to align more quickly.

    • The correct option is D.

    20. When whirling a stone in a horizontal circle, the angular velocity is
    Angular velocity is a vector quantity whose direction is along the axis of rotation. For a stone whirled in a horizontal circle, the axis of rotation is vertical, meaning the angular velocity vector is perpendicular to the plane of the circle.

    • The correct option is B.

    **21. Which of the following therm

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