Physics-
General
Easy

Question

The figure shows a system of two concentric spheres of radii r subscript 1 end subscript and r subscript 2 end subscript and kept at temperatures T subscript 1 end subscript and T subscript 2 end subscript respectively. The radial rate of flow of heat in a substance between the two concentric spheres, is proportional to

  1. fraction numerator left parenthesis r subscript 2 end subscript minus r subscript 1 end subscript right parenthesis over denominator left parenthesis r subscript 1 end subscript r subscript 2 end subscript right parenthesis end fraction    
  2. Inopen parentheses fraction numerator r subscript 2 end subscript over denominator r subscript 1 end subscript end fraction close parentheses    
  3. fraction numerator open parentheses r subscript 1 end subscript r subscript 2 end subscript close parentheses over denominator open parentheses r subscript 2 end subscript minus r subscript 1 end subscript close parentheses end fraction    
  4. left parenthesis r subscript 2 end subscript minus r subscript 1 end subscript right parenthesis    

The correct answer is: fraction numerator open parentheses r subscript 1 end subscript r subscript 2 end subscript close parentheses over denominator open parentheses r subscript 2 end subscript minus r subscript 1 end subscript close parentheses end fraction


    To measure the radial rate of heat flow, we have to go for integration technique as here the area of the surface through which heat will flow is not constant.

    Let us consider an element (spherical shell) of thickness dx and radius x as shown in figure. Let us first find the equivalent thermal resistance between inner and outer sphere.
    Resistance of shell=d R equals blank fraction numerator d x over denominator K cross times 4 pi x to the power of 2 end exponent end fraction
    open parentheses fraction numerator F r o m blank R equals fraction numerator l over denominator K A end fraction w h e r e comma over denominator K equals t h e r m a l blank c o n d u c t i v i t y end fraction close parentheses
    rightwards double arrow not stretchy integral d R equals R equals not stretchy integral subscript r subscript 1 end subscript end subscript superscript r subscript 2 end subscript end superscript fraction numerator d x over denominator 4 pi K x to the power of 2 blank end exponent end fraction
    = fraction numerator 1 over denominator 4 pi K end fraction open square brackets fraction numerator 1 over denominator r subscript 1 end subscript end fraction minus blank fraction numerator 1 over denominator r subscript 2 end subscript end fraction close square brackets equals blank fraction numerator r subscript 2 end subscript minus r subscript 1 end subscript over denominator 4 pi K left parenthesis r subscript 1 end subscript r subscript 2 end subscript right parenthesis end fraction
    Rate of heat flow = H
    = fraction numerator T subscript 1 end subscript minus T subscript 2 end subscript over denominator R end fraction
    = fraction numerator T subscript 1 end subscript minus T subscript 2 end subscript over denominator r subscript 2 end subscript minus r subscript 1 end subscript end fraction blank cross times 4 pi K open parentheses r subscript 1 end subscript r subscript 2 end subscript close parentheses
    proportional to fraction numerator r subscript 1 end subscript r subscript 2 end subscript over denominator r subscript 2 end subscript minus blank r subscript 1 end subscript end fraction

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    If the line y = x + 3 meets the circle x squared plus y squared equals a squared at A and B, then equation of the circle on AB as diameter is

    If the line y = x + 3 meets the circle x squared plus y squared equals a squared at A and B, then equation of the circle on AB as diameter is

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    An electric lamp is fixed at the ceiling of a circular tunnel as shown is figure. What is the ratio the intensities of light at base A and a point B on the wall

    I subscript A end subscript equals fraction numerator L over denominator left parenthesis 2 r right parenthesis to the power of 2 end exponent end fraction and I subscript B end subscript equals fraction numerator L over denominator open parentheses r square root of 2 close parentheses to the power of 2 end exponent end fraction cos invisible function application theta
    equals fraction numerator L over denominator 2 r to the power of 2 end exponent end fraction. fraction numerator r over denominator r square root of 2 end fraction equals fraction numerator L over denominator 2 square root of 2   r to the power of 2 end exponent end fraction

    therefore fraction numerator I subscript A end subscript over denominator I subscript B end subscript end fraction equals fraction numerator 2 square root of 2 over denominator 4 end fraction equals fraction numerator 1 over denominator square root of 2 end fraction

    An electric lamp is fixed at the ceiling of a circular tunnel as shown is figure. What is the ratio the intensities of light at base A and a point B on the wall

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    I subscript A end subscript equals fraction numerator L over denominator left parenthesis 2 r right parenthesis to the power of 2 end exponent end fraction and I subscript B end subscript equals fraction numerator L over denominator open parentheses r square root of 2 close parentheses to the power of 2 end exponent end fraction cos invisible function application theta
    equals fraction numerator L over denominator 2 r to the power of 2 end exponent end fraction. fraction numerator r over denominator r square root of 2 end fraction equals fraction numerator L over denominator 2 square root of 2   r to the power of 2 end exponent end fraction

    therefore fraction numerator I subscript A end subscript over denominator I subscript B end subscript end fraction equals fraction numerator 2 square root of 2 over denominator 4 end fraction equals fraction numerator 1 over denominator square root of 2 end fraction
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    The distance between a point source of light and a screen which is 60 cm is increased to 180 cm. The intensity on the screen as compared with the original intensity will be

    I proportional to fraction numerator 1 over denominator r to the power of 2 end exponent end fraction rightwards double arrow fraction numerator I subscript 2 end subscript over denominator I subscript 1 end subscript end fraction equals fraction numerator r subscript 1 end subscript superscript 2 end superscript over denominator r subscript 2 end subscript superscript 2 end superscript end fraction equals fraction numerator 6 0 to the power of 2 end exponent over denominator 18 0 to the power of 2 end exponent end fraction equals fraction numerator 1 over denominator 9 end fraction

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    I proportional to fraction numerator 1 over denominator r to the power of 2 end exponent end fraction rightwards double arrow fraction numerator I subscript 2 end subscript over denominator I subscript 1 end subscript end fraction equals fraction numerator r subscript 1 end subscript superscript 2 end superscript over denominator r subscript 2 end subscript superscript 2 end superscript end fraction equals fraction numerator 6 0 to the power of 2 end exponent over denominator 18 0 to the power of 2 end exponent end fraction equals fraction numerator 1 over denominator 9 end fraction
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    The pole of the straight line 9x + y – 28 = 0 with respect to the circle 2 x squared plus 2 y squared minus 3 x plus 5 y minus 7 equals 0 is

    Let the coordinates of the pole be (h, k).
    The equation of the circle is 2x2 + 2y2 - 3x + 5y - 7 = 0.
    Dividing both sides of the equation by 2,we get
    x2 + y2 - 3/2 x + 5/2 y - 7/2 = 0.
    The equation of polar for the point having coordinates (h, k) is
    h x + k y - 3/4 (x + h) + 5/4 (y+k) - 7/2 = 0 .
    Simplifying the equation, it can be written as (4h - 3)x + (4k + 5)y + (-3h + 5k - 14).
    We are given that the equation of the polar is 9x + y - 28 = 0.so, the above equation and this equation represent the same line. Comparing the coefficient of the line, we get
    (4h - 3) /9 = (4k + 5) /1 = (-3h + 5k - 14) /-28
    Taking the first and third expression, we get
    (4h - 3) /9 = (-3h + 5k - 14) /-28 .
    By cross multiplication, we get
    Or, -28 (4h - 3) = 9 (-3h + 5k - 14)
    Or, -112h + 84 = -27h + 45k - 126
    Or, -112h + 27h - 45k = -126 - 84
    Or, -85h - 45k = -210
    Or, -5 ( 17h + 9k ) = -210
    17h + 9k = 42 . . . (1)
    Taking the second and third expression, we get
    (4k + 5) /1 = ( -3h + 5k - 14) /-28
    By cross multiplication, we get
    -28 ( 4k + 5 ) = ( -3h + 5k - 14 )
    -112k - 140 = -3h + 5k - 14
    3h - 112k - 5k = -14 + 140
    3h - 117k = 126 . . . (2)
    By solving equation (1) × 13 and equation (2) .
    h = 3
    By putting the value of h in equation (1) , we get
    K = -1 .
    Therefore, the coordinates are (3, -1) .
    Hence, the correct option is (a).

    The pole of the straight line 9x + y – 28 = 0 with respect to the circle 2 x squared plus 2 y squared minus 3 x plus 5 y minus 7 equals 0 is

    Maths-General
    Let the coordinates of the pole be (h, k).
    The equation of the circle is 2x2 + 2y2 - 3x + 5y - 7 = 0.
    Dividing both sides of the equation by 2,we get
    x2 + y2 - 3/2 x + 5/2 y - 7/2 = 0.
    The equation of polar for the point having coordinates (h, k) is
    h x + k y - 3/4 (x + h) + 5/4 (y+k) - 7/2 = 0 .
    Simplifying the equation, it can be written as (4h - 3)x + (4k + 5)y + (-3h + 5k - 14).
    We are given that the equation of the polar is 9x + y - 28 = 0.so, the above equation and this equation represent the same line. Comparing the coefficient of the line, we get
    (4h - 3) /9 = (4k + 5) /1 = (-3h + 5k - 14) /-28
    Taking the first and third expression, we get
    (4h - 3) /9 = (-3h + 5k - 14) /-28 .
    By cross multiplication, we get
    Or, -28 (4h - 3) = 9 (-3h + 5k - 14)
    Or, -112h + 84 = -27h + 45k - 126
    Or, -112h + 27h - 45k = -126 - 84
    Or, -85h - 45k = -210
    Or, -5 ( 17h + 9k ) = -210
    17h + 9k = 42 . . . (1)
    Taking the second and third expression, we get
    (4k + 5) /1 = ( -3h + 5k - 14) /-28
    By cross multiplication, we get
    -28 ( 4k + 5 ) = ( -3h + 5k - 14 )
    -112k - 140 = -3h + 5k - 14
    3h - 112k - 5k = -14 + 140
    3h - 117k = 126 . . . (2)
    By solving equation (1) × 13 and equation (2) .
    h = 3
    By putting the value of h in equation (1) , we get
    K = -1 .
    Therefore, the coordinates are (3, -1) .
    Hence, the correct option is (a).
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    Three rods of equal length l are joined to form an equilateral triangle P Q R. O is the mid point of P Q. Distance O R remains same for small change in temperature. Coefficient of linear expansion for P R and R Q is same, i. e. comma blank alpha subscript 2 end subscript but that for P Q is alpha subscript 1 end subscript. Then

    open parentheses O R close parentheses to the power of 2 end exponent equals open parentheses P R close parentheses to the power of 2 end exponent minus open parentheses P O close parentheses to the power of 2 end exponent equals l to the power of 2 end exponent minus open parentheses fraction numerator l over denominator 2 end fraction close parentheses to the power of 2 end exponent
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    open parentheses O R close parentheses to the power of 2 end exponent equals open parentheses P R close parentheses to the power of 2 end exponent minus open parentheses P O close parentheses to the power of 2 end exponent equals l to the power of 2 end exponent minus open parentheses fraction numerator l over denominator 2 end fraction close parentheses to the power of 2 end exponent
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    Neglecting alpha subscript 2 end subscript superscript 2 end superscript t to the power of 2 end exponent and alpha subscript 1 end subscript superscript 2 end superscript t to the power of 2 end exponent
    0 equals l to the power of 2 end exponent open parentheses 2 alpha subscript 2 end subscript t close parentheses minus fraction numerator l to the power of 2 end exponent over denominator 4 end fraction open parentheses 2 alpha subscript 1 end subscript t close parentheses rightwards double arrow 2 alpha subscript 2 end subscript equals fraction numerator 2 alpha subscript 1 end subscript over denominator 4 end fraction rightwards double arrow alpha subscript 1 end subscript equals 4 alpha subscript 2 end subscript
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    A point source causes photoelectric effect from a small metal plate Which of the following curves may represent the saturation photocurrent as a function of the distance between the source and the metal?

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    One of the following figures respesents the variation of particle momentum with associated de Broglie wavelength
    a)
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    One of the following figures respesents the variation of particle momentum with associated de Broglie wavelength
    a)
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    Two circular discs A and B with equal radii are blackened. They are heated to some temperature and are cooled under identical conditions. What inference do you draw from their cooling curves?

    According to Newton’s law of cooling, rate of cooling is given by
    open parentheses fraction numerator negative d T over denominator d t end fraction close parentheses equals fraction numerator e A sigma over denominator m c end fraction left parenthesis T to the power of 4 end exponent minus T subscript 0 end subscript superscript 4 end superscript right parenthesis
    Where c is specific heat of material.
    or open parentheses fraction numerator negative d T over denominator d t end fraction close parentheses proportional to fraction numerator 1 over denominator c end fraction
    i e. comma blankrate of cooling varies inversely as specific heat. From the graph, for A rate of cooling is larger. Therefore, specific heat of A is smaller.

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    open parentheses fraction numerator negative d T over denominator d t end fraction close parentheses equals fraction numerator e A sigma over denominator m c end fraction left parenthesis T to the power of 4 end exponent minus T subscript 0 end subscript superscript 4 end superscript right parenthesis
    Where c is specific heat of material.
    or open parentheses fraction numerator negative d T over denominator d t end fraction close parentheses proportional to fraction numerator 1 over denominator c end fraction
    i e. comma blankrate of cooling varies inversely as specific heat. From the graph, for A rate of cooling is larger. Therefore, specific heat of A is smaller.
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    Which of the curves in figure represents the relation between Celsius and Fahrenheit temperatures

    fraction numerator C over denominator 5 end fraction equals fraction numerator F minus 32 over denominator 9 end fraction rightwards double arrow C equals open parentheses fraction numerator 5 over denominator 9 end fraction close parentheses F minus fraction numerator 20 over denominator 3 end fraction. Hence graph between ℃ and ℉ will be a straight line with positive slope and negative intercept

    Which of the curves in figure represents the relation between Celsius and Fahrenheit temperatures

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    fraction numerator C over denominator 5 end fraction equals fraction numerator F minus 32 over denominator 9 end fraction rightwards double arrow C equals open parentheses fraction numerator 5 over denominator 9 end fraction close parentheses F minus fraction numerator 20 over denominator 3 end fraction. Hence graph between ℃ and ℉ will be a straight line with positive slope and negative intercept
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    The area of circle centred at (1, 2) and passing through (4, 6) is

    Here, we have to find the area of circle.
    Now, the centre of circle is C (1, 2).
    The circle passes through the point P(4, 6).
    Radius = CP
    = √(4-1) 2 + (6-2) 2
    = √(3) 2 + (4) 2
    = √9+16
    =√25
    =5
    Therefore, radius is 5 unit.
    Now, Area of circle = π × 5 ×5
    = 25π .
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    The area of circle centred at (1, 2) and passing through (4, 6) is

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    Here, we have to find the area of circle.
    Now, the centre of circle is C (1, 2).
    The circle passes through the point P(4, 6).
    Radius = CP
    = √(4-1) 2 + (6-2) 2
    = √(3) 2 + (4) 2
    = √9+16
    =√25
    =5
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    Now, Area of circle = π × 5 ×5
    = 25π .
    Hence, the correct option is (a).
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    Three rods of same dimensions are arranged as shown in figure. They have thermal conductivities K subscript 1 end subscript comma K subscript 2 end subscript and K subscript 3 end subscript. The points P and Q are maintained at different temperatures for the heat to flow at the same rate along P R Q and P Q then which of the following options is correct

    The given arrangement of rods can be redrawn as follows

    It is given that H subscript 1 end subscript equals H subscript 2 end subscript
    rightwards double arrow fraction numerator K A left parenthesis theta subscript 1 end subscript minus theta subscript 2 end subscript right parenthesis over denominator 2 l end fraction equals fraction numerator K subscript 3 end subscript A left parenthesis theta subscript 1 end subscript minus theta subscript 2 end subscript right parenthesis over denominator l end fraction rightwards double arrow K subscript 3 end subscript equals fraction numerator K over denominator 2 end fraction equals fraction numerator K subscript 1 end subscript K subscript 2 end subscript over denominator K subscript 1 end subscript plus K subscript 2 end subscript end fraction

    Three rods of same dimensions are arranged as shown in figure. They have thermal conductivities K subscript 1 end subscript comma K subscript 2 end subscript and K subscript 3 end subscript. The points P and Q are maintained at different temperatures for the heat to flow at the same rate along P R Q and P Q then which of the following options is correct

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    The given arrangement of rods can be redrawn as follows

    It is given that H subscript 1 end subscript equals H subscript 2 end subscript
    rightwards double arrow fraction numerator K A left parenthesis theta subscript 1 end subscript minus theta subscript 2 end subscript right parenthesis over denominator 2 l end fraction equals fraction numerator K subscript 3 end subscript A left parenthesis theta subscript 1 end subscript minus theta subscript 2 end subscript right parenthesis over denominator l end fraction rightwards double arrow K subscript 3 end subscript equals fraction numerator K over denominator 2 end fraction equals fraction numerator K subscript 1 end subscript K subscript 2 end subscript over denominator K subscript 1 end subscript plus K subscript 2 end subscript end fraction
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    In the following diagram if V subscript 2 end subscript greater than V subscript 1 end subscript then 

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    The maximum kinetic energy (Ek ) of emitted photoelectrons against frequency v of incident radiation is plotted as shown in fig The slope of the graph is equal to

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    A lamp rated at 100 cd hangs over the middle of a round table with diameter 3 m at a height of 2 m. It is replaced by a lamp of 25 cd and the distance to the table is changed so that the illumination at the centre of the table remains as before. The illumination at edge of the table becomes X times the original. Then X is

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