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JEE Advanced Mix Test 11

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JEE Advanced Mix Test 11
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Weekly Quiz Competition
  • Question 1
    4 / -1

    4 gms of steam at 100°C is added to 20 gms of water at 46°C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-°C.

    Solution

    Heat released by steam inconversion to water at 100°C is Q1 = mL = 4 × 540 = 2160 cal.

    Heat required to raise temperature of water from 46°C t 100°C is Q2 = mS Δθ = 20 × 1 × 54 = 1080 J

    Hence all steam is not converted to water only half steam shall be converted to water

    So Final mass of water = 20 + 2 = 22 gm

  • Question 2
    4 / -1

    A diatomic ideal gas undergoes a thermodynamic change according to the P–V diagram shown in the figure. The total heat given to the gas is nearly (use ln2 = 0.7) :

    Solution

  • Question 3
    4 / -1

    Two rods are joined between fixed supports as shown in the figure. Condition for no change in the lengths of individual rods with the increase of temperature will be ( α1, α2 = linear expansion co-efficient A1, A2 = Area of rods Y1, Y2 = Young modulus).

    Solution

  • Question 4
    4 / -1

    Solution

  • Question 5
    4 / -1

    The co-efficient of thermal expansion of a rod is temperature dependent and is given by the formula α = a T, where a is a positive constant and T in ºC. If the length of the rod is l at temperature 0 ºC, then the temperature at which the length will be 2 l is:

    Solution

  • Question 6
    4 / -1

    A black body emits radiation at the rate P when its temperature is T. At this temperature the wavelength at which the radiation has maximum intensity is λ0. If at another temperature T' the power radiated is

    Solution

  • Question 7
    4 / -1

    Thermal coefficient of volume expansion at constant pressure for an ideal gas sample of n moles having pressure P0, volume V0 and temperature T0 is

    Solution

  • Question 8
    4 / -1

    There are two thin spheres A and B of the same material and same thickness. They emit like black bodies. Radius of A is double that of B. A and B have same temperature T. When A and B are kept in a room of temperature T0 (< T), the ratio of their rates of cooling (rate of fall of temperature) is: [ assume negligible heat exchange between A and B ]

    Solution

  • Question 9
    4 / -1

    Solution

  • Question 10
    4 / -1

    When the temperature of a copper coin is raised by 80 oC, its diameter increases by 0.2%,

    Solution

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