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Physics Test - 5

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Physics Test - 5
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  • Question 1
    3 / -1

    A child running at a temperature of 101oF is given an antipyrin (i.e., medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98oF in 20 min, what is the average rate of extra evaporation caused by which heat is lost. The mass of child is 30 kg. The specific heat of the human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about 580 cal/g.

    Solution

    Given, mass of the child (m) = 30 kg

    Time taken (t)  = 20 min

    Fall in temperature = (101 - 98)oF

    Specific heat of human body (s) = 4.2 x 103 J/kg - oC

    Latenl heat of evaporation (L) = 580 cal/g = 580 x 103 cal/kg

    = (580 x 103 x 4.2) J/kg

    Heat given by body during fall in temperature

    Q1 = ms ΔT

    Let m' be the mass of sweat evaporates from the human body.

    Heat taken in evaporation

    Q2 = m' L

    But  Q1 = Q2

    ∴ msΔT = m′L

    or

  • Question 2
    3 / -1

    A sphere of radius 10 cm is hung inside an oven whose walls are at a temperature of 1000 K. Calculate the total heat energy incident per second on the sphere. Given that Stefan's constant, σ = 5.67 × 10−8 W m−1 °C−4.

    Solution

    Here, r = 10 cm = 0.1 m

    T = 1000 K

    σ = 5.67 × 10−8 W m−1 °C−4.

    If A is the surface area of the sphere then, heat energy incident per second on the sphere,

    E = A × (σT4) = 4πr× σT4

    = 4π × (0.1)× 5.67 × 10−8 × (1,000)= 7,125 J s−1  

  • Question 3
    3 / -1

    The thermal conductivity of a rod is 2. What is its thermal resistivity?

    Solution

  • Question 4
    3 / -1

    The black body spectrum of an object Q1 is such that its radiant intensity (i.e., intensity per unit wavelength interval) is maximum at a wavelength of 100 nm. Another object O2 has the maximum radiant intensity at 600 nm. The ratio of power emitted per unit area by source O1 to that of source O2 is

    Solution

    From Wein's displacement law,

  • Question 5
    3 / -1

    Assertion: In a process if initial volume is equal to the final volume, work done by the gas is zero.

    Reason: In an isochoric process work done by the gas is zero.

    Solution

    Work done in a process will be zero if the volume remains constant throughout the process(Isochoric process). If the volume changes in a process, the work done in the process will not be zero even if the final volume and the initial volume are equal(Example: Vertical semi-circle in a PV graph).

    Therefore, the Assertion is false but Reason is true.

  • Question 6
    3 / -1

    The ratio of thermal capacities of two spheres A and B, if their diameters are in the ratio 1:2, densities in the ratio 2:1, and the specific heat in the ratio of 1:3, will be

    Solution

  • Question 7
    3 / -1

    The rate of emission of radiation of a black body at temperature 27℃ is E1. If its temperature is increased to 327℃, the rate of emission of radiation is E2. The relation between E1 and E2 is

    Solution

    Here, T1 = 27℃ = (27 + 273)K = 300 K

    T2 = 327℃ (327 + 273)K = 600 K

    According to Stefan's law,

    E = eAδT4

  • Question 8
    3 / -1

    Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the ratio of TA and TB ?

    Solution

    Given, triple point of water on scale A = 200 A

    Triple point of water on scale B = 350 B

    We know that triple point of water on absolute scale = 273.16 K

    ∴ 200 A = 350 B = 273.16 K

  • Question 9
    3 / -1

    Following graph shows the correct variation in intensity of heat radiations by black body and frequency at a fixed temperature

    Solution

    According to Wien's law, 

    As the temperature of body increases, frequency corresponding to maximum energy in radiation (vm) increases. This is shown in graph (C).

  • Question 10
    3 / -1

    The thermodynamic state of two moles of an ideal gas is changed from A to B along the path shown. The net work done by the gas is

    Solution

    Work done by a gas is equal to the area under the curve on volume axis.

    Since cycle is anticlockwise so negative work done is more than the positive work done. S o overall work done will be negative.

    Hence option (B) is the correct answer of the given diagram.

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