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Physics Test 211

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Physics Test 211
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  • Question 1
    4 / -1

    A block of mass 0.5 kg is at rest on a horizontal table. The coefficient of kinetic friction between the table and the block is 0.2. If a horizontal force of 5 N is applied on the block, the kinetic energy of the block in a time of 4 s is (Acceleration due to gravity = 10ms−2

    Solution

    Given values:

    • Mass of the block (m) = 0.5 kg

    • Applied force (Fapplied) = 5 N

    • Coefficient of kinetic friction (μk) = 0.2

    • Acceleration due to gravity (g) = 10 m/s2

    • Time (t) = 4 s

    The frictional force (ffriction):

    ffriction = μk × m × g

    ffriction = 0.2 × 0.5 kg × 10 m/s2 = 1 N

    The net force (Fnet):

    Fnet = Fapplied − ffriction

    Fnet = 5 N − 1 N = 4 N

    The acceleration (a):

    a = Fnet / m

    a = 4 N / 0.5 kg = 8 m/s2

    The velocity (v) after 4 seconds:

    v = u + a × t

    v = 0 + 8 m/s2 × 4 s = 32 m/s

    The kinetic energy (KE):

    KE = (1/2) × m × v2

    KE = (1/2) × 0.5 kg × (32 m/s)2 = (1/2) × 0.5 × 1024 = 256 J

    The kinetic energy of the block after 4 seconds is 256 J.

     

  • Question 2
    4 / -1

    The potential energy of a particle (Ux) executing S.H.M. is given by

    Solution

    Given the potential energy function, we compare it with the standard form of potential energy in S.H.M.

    Option 1: Ux = (k/2)(x - a)2

    ⇒ This matches the standard form Ux = (1/2) k (x - a)2

    ∴ The correct answer is Option 1: Ux = (k/2)(x - a)2.

     

  • Question 3
    4 / -1

    A system consists of three particles each of mass 'm1' placed at the corners of an equilateral triangle of side  A particle of mass 'm2' is placed at the mid point of any one side of the triangle. Due to the system of particles, the force acting on m2 is

    Solution

    Identify Distances

    The distance from each corner to the midpoint of the side is L/6.

    Calculate Force from One Corner

    The force exerted by corner A and Corner C will cancel out as they are at equal distance from mass m1 is:

    Net Force from All Three Masses:

    FA+FB+FC=FB

    Total force:

    Ftotal = FB= 12G × (m1 × m2) / h2

    = G × (m1 × m2) / ((√3 /2)× (L/3))2

    = 12G × (m1 × m2) / L2

    ∴ The correct answer is 12G × (m1 × m2) / L2.

     

  • Question 4
    4 / -1

    The stress versus strain graphs for wires of two materials A and B are as shown in the figure. If YA and YB are the Young's modulii of the materials, then:

    Solution

    Stress/Strain = Y = slope of graph

    ∴YA /YB

    = tan(60°) / tan(30°)

    = 3 ∴ YA = 3YB

     

  • Question 5
    4 / -1

    Four massless springs whose force constants are 2k, 2k, k and 2k respectively are attached to a mass M kept on a frictionless plane (as shown in figure). If the mass M is displaced in the horizontal direction, then the frequency of the system

    Solution

    The frequency of oscillation is given by:

    ν = (1/2π) √(K/M)

    ⇒ ν = (1/2π) √(4k/M)

    ∴ The required frequency is (1/2π) √(4k/M).

     

  • Question 6
    4 / -1

    A steel wire of length 3 m and a copper wire of length 2.2 m are connected end to end. When the combination is stretched by a force, the net elongation is 1.05 mm. If the area of cross-section of each wire is 6mm2, then the load applied is (Young's moduli of steel and copper are respectively 2 × 1011 Nm-2 and 1.1 × 1011 Nm-2 ).

    Solution

    Given,

    Length of steel wire, Lsteel = 3 m

    Length of copper wire, Lcopper = 2.2 m

    Area of cross-section of each wire, A = 6 mm2 = 6 × 10-6 m2

    Elongation of the system, ΔLtotal = 1.05 mm = 1.05 × 10-3 m

    You are required to find the load applied (F). For this, we use the total elongation formula:

    ΔLtotal = ΔLsteel + ΔLcopper

    For steel wire, elongation is: ΔLsteel = F Lsteel / (A Ysteel)

    For copper wire, elongation is: ΔLcopper = F Lcopper / (A Ycopper)

    Thus, the total elongation is:

    ΔLtotal = F Lsteel / (A Ysteel) + F Lcopper / (A Ycopper)

    Substitute the known values:

    1.05 × 10-3 = F (3 / (6 × 10-6 × 2 × 1011) + 2.2 / (6 × 10-6 × 1.1 × 1011))

    Solving the above equation:

    1.05 × 10-3 = F (3 / 1.2 × 106 + 2.2 / 6.6 × 105)

    1.05 × 10-3 = F (2.5 × 10-6 + 3.33 × 10-6)

    1.05 × 10-3 = F × 5.83 × 10-6

    F = (1.05 × 10-3) / (5.83 × 10-6) = 180.9 N

    ∴ The load applied is approximately 180 N.

    Hence, correct option is 1) 180 N.

     

  • Question 7
    4 / -1

    Solution

    Correct Option: 1

    Explanation:

    • Thermal conductivity: The dimensional formula for thermal conductivity is MLT-3K-1. It relates heat transfer per unit area to the temperature gradient.

    • Boltzmann constant: The dimensional formula for the Boltzmann constant is M0L2T-2K-1. It is a fundamental constant connecting temperature with energy.

    • Latent heat: The dimensional formula for latent heat is ML2T-2. It represents the heat required for a phase change per unit mass.

    • Specific heat: The dimensional formula for specific heat is M0L2T-2. It defines the heat required to raise the temperature of a unit mass by one degree.

    Matching:

    • (a) Thermal conductivity → (i) MLT-3K-1

    • (b) Boltzmann constant → (iii) ML2T-2K-1

    • (c) Latent heat → (iv) M0L2T-2

    • (d) Specific heat → (ii) M0L2T-2K-1

    Conclusion:

    ∴ The correct matching is: a - i, b - iii, c - iv, d - ii.

     

  • Question 8
    4 / -1

    A ball of mass 0.2kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2m while applying the force and the ball goes upto 2m height further, find the magnitude of the force. (Consider g = 10m/s2).

    Solution

    Given: m = 0.2 kg, g = 10 m/s², h = 2 m, distance = 0.2 m.

    PE = 0.2 × 10 × 2 = 4 J.

    Work = F × 0.2, so F = 4 / 0.2 = 20 N.

    Adding weight: 20 N + 2 N = 22 N.

    ∴ The force applied is 22 N. Option 4 is correct.

     

  • Question 9
    4 / -1

    In the following diagram, the work done in moving a point charge from point P to point A, B and C are WA, WB, and WC respectively. Then (A, B, C are points on semicircle and point charge q is at the centre of semicircle)

    Solution

    Given:

    Points A, B, and C are equidistant from charge q at the center.

    ⇒ VA = VB = VC

    ⇒ Work done in moving a charge from P to A, B, or C is the same.

    ⇒ WA = WB = WC

    Since moving from P to A, B, or C requires work, the work done is not zero.

    ∴ The correct answer is WA = WB = WC ≠ 0.

     

  • Question 10
    4 / -1

    A particle is projected with velocity v0 along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma = -αx2. The distance at which the particle stops:

    Solution

    Thus, most suitable answer could be (3) as mass ‘m’ is not given in any options.

    ∴ The correct option is 3

     

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