Maths-
General
Easy

Question

If s i n invisible function application x equals 1 divided by 2, then s i n invisible function application 3 x equals_______________

  1. square root of 3 divided by 2
  2. 1 divided by 2    
  3. 1 divided by square root of 2    
  4. 1    

The correct answer is: square root of 3 divided by 2

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Related Questions to study

General
physics-

Which one of the following is v subscript m end subscript minus T graph for perfectly black body? v subscript m end subscriptis the frequency of radiation with maximum intensity, T is the absolute temperature.

Intensity is directly proportional to energy.

Which one of the following is v subscript m end subscript minus T graph for perfectly black body? v subscript m end subscriptis the frequency of radiation with maximum intensity, T is the absolute temperature.

physics-General
Intensity is directly proportional to energy.
General
physics-

Three rods made of same material and having same cross-section are joined as shown in the figure. Each rod is of same length. The temperature at the junction of the three rods is

Let the temperature of function be theta, then
H equals H subscript 1 end subscript plus H subscript 2 end subscript
rightwards double arrow fraction numerator K A left parenthesis theta minus 0 right parenthesis over denominator L end fraction equals fraction numerator K A left parenthesis 90 minus theta right parenthesis over denominator L end fraction plus fraction numerator K A left parenthesis 90 minus theta right parenthesis over denominator L end fraction

Or theta equals 90 minus theta plus 90 minus theta
Or theta equals 180 minus 2 theta
Or 3 theta equals 180
Or theta equals 60 ℃

Three rods made of same material and having same cross-section are joined as shown in the figure. Each rod is of same length. The temperature at the junction of the three rods is

physics-General
Let the temperature of function be theta, then
H equals H subscript 1 end subscript plus H subscript 2 end subscript
rightwards double arrow fraction numerator K A left parenthesis theta minus 0 right parenthesis over denominator L end fraction equals fraction numerator K A left parenthesis 90 minus theta right parenthesis over denominator L end fraction plus fraction numerator K A left parenthesis 90 minus theta right parenthesis over denominator L end fraction

Or theta equals 90 minus theta plus 90 minus theta
Or theta equals 180 minus 2 theta
Or 3 theta equals 180
Or theta equals 60 ℃
General
physics-

The figure shows a glass tube (linear co-efficient of expansion is alpha) completely filled with a liquid of volume expansion co-efficient gamma. On heating length of the liquid column does not change. Choose the correct relation between gamma and alpha

When length of the liquid column remains constant, then the level of liquid moves down with respect to the container, thus gamma must be less than 3 alpha
Now we can write V equals V subscript 0 end subscript left parenthesis 1 plus gamma increment T right parenthesis
Since V equals A l subscript 0 end subscript equals open square brackets A subscript 0 end subscript open parentheses 1 plus 2 alpha increment T close parentheses close square brackets l subscript 0 end subscript equals V subscript 0 end subscript open parentheses 1 plus 2 alpha increment T close parentheses
Hence V subscript 0 end subscript open parentheses 1 plus gamma increment T close parentheses equals V subscript 0 end subscript open parentheses 1 plus 2 alpha increment T close parentheses rightwards double arrow gamma equals 2 alpha

The figure shows a glass tube (linear co-efficient of expansion is alpha) completely filled with a liquid of volume expansion co-efficient gamma. On heating length of the liquid column does not change. Choose the correct relation between gamma and alpha

physics-General
When length of the liquid column remains constant, then the level of liquid moves down with respect to the container, thus gamma must be less than 3 alpha
Now we can write V equals V subscript 0 end subscript left parenthesis 1 plus gamma increment T right parenthesis
Since V equals A l subscript 0 end subscript equals open square brackets A subscript 0 end subscript open parentheses 1 plus 2 alpha increment T close parentheses close square brackets l subscript 0 end subscript equals V subscript 0 end subscript open parentheses 1 plus 2 alpha increment T close parentheses
Hence V subscript 0 end subscript open parentheses 1 plus gamma increment T close parentheses equals V subscript 0 end subscript open parentheses 1 plus 2 alpha increment T close parentheses rightwards double arrow gamma equals 2 alpha
General
physics-

The spectrum of a black body at two temperatures 27 ℃ and 327 ℃ is shown in the figure. Let A subscript 1 end subscript and A subscript 2 end subscript be the areas under the two curves respectively. The value of fraction numerator A subscript 2 end subscript over denominator A subscript 1 end subscript end fraction is

Area under given curve represents emissive power and emissive power proportional to T to the power of 4 end exponent rightwards double arrow A proportional to T to the power of 4 end exponent
rightwards double arrow fraction numerator A subscript 2 end subscript over denominator A subscript 1 end subscript end fraction equals fraction numerator T subscript 2 end subscript superscript 4 end superscript over denominator T subscript 1 end subscript superscript 4 end superscript end fraction equals fraction numerator open parentheses 273 plus 327 close parentheses to the power of 4 end exponent over denominator open parentheses 273 plus 27 close parentheses to the power of 4 end exponent end fraction equals open parentheses fraction numerator 600 over denominator 300 end fraction close parentheses to the power of 4 end exponent equals fraction numerator 16 over denominator 1 end fraction

The spectrum of a black body at two temperatures 27 ℃ and 327 ℃ is shown in the figure. Let A subscript 1 end subscript and A subscript 2 end subscript be the areas under the two curves respectively. The value of fraction numerator A subscript 2 end subscript over denominator A subscript 1 end subscript end fraction is

physics-General
Area under given curve represents emissive power and emissive power proportional to T to the power of 4 end exponent rightwards double arrow A proportional to T to the power of 4 end exponent
rightwards double arrow fraction numerator A subscript 2 end subscript over denominator A subscript 1 end subscript end fraction equals fraction numerator T subscript 2 end subscript superscript 4 end superscript over denominator T subscript 1 end subscript superscript 4 end superscript end fraction equals fraction numerator open parentheses 273 plus 327 close parentheses to the power of 4 end exponent over denominator open parentheses 273 plus 27 close parentheses to the power of 4 end exponent end fraction equals open parentheses fraction numerator 600 over denominator 300 end fraction close parentheses to the power of 4 end exponent equals fraction numerator 16 over denominator 1 end fraction
General
physics-

In a vertical U-tube containing a liquid, the two arms are maintained at different temperatures t subscript 1 end subscript and t subscript 2 end subscript. The liquid columns in the two arms have heights l subscript 1 end subscript and l subscript 2 end subscript respectively. The coefficient of volume expansion of the liquid is equal to

Suppose, height of liquid in each arm before rising the temperature is l.

With temperature rise height of liquid in each arm increases i. e. blank l subscript 1 end subscript greater than l blank a n d blank l subscript 2 end subscript greater than l
Also l equals fraction numerator l subscript 1 end subscript over denominator 1 plus gamma t subscript 1 end subscript end fraction equals fraction numerator l subscript 2 end subscript over denominator 1 plus gamma t subscript 2 end subscript end fraction
rightwards double arrow l subscript 1 end subscript plus gamma l subscript 1 end subscript t subscript 2 end subscript equals l subscript 2 end subscript plus gamma l subscript 2 end subscript t subscript 1 end subscript rightwards double arrow gamma equals fraction numerator l subscript 1 end subscript minus l subscript 2 end subscript over denominator l subscript 2 end subscript t subscript 1 end subscript minus l subscript 1 end subscript t subscript 2 end subscript end fraction

In a vertical U-tube containing a liquid, the two arms are maintained at different temperatures t subscript 1 end subscript and t subscript 2 end subscript. The liquid columns in the two arms have heights l subscript 1 end subscript and l subscript 2 end subscript respectively. The coefficient of volume expansion of the liquid is equal to

physics-General
Suppose, height of liquid in each arm before rising the temperature is l.

With temperature rise height of liquid in each arm increases i. e. blank l subscript 1 end subscript greater than l blank a n d blank l subscript 2 end subscript greater than l
Also l equals fraction numerator l subscript 1 end subscript over denominator 1 plus gamma t subscript 1 end subscript end fraction equals fraction numerator l subscript 2 end subscript over denominator 1 plus gamma t subscript 2 end subscript end fraction
rightwards double arrow l subscript 1 end subscript plus gamma l subscript 1 end subscript t subscript 2 end subscript equals l subscript 2 end subscript plus gamma l subscript 2 end subscript t subscript 1 end subscript rightwards double arrow gamma equals fraction numerator l subscript 1 end subscript minus l subscript 2 end subscript over denominator l subscript 2 end subscript t subscript 1 end subscript minus l subscript 1 end subscript t subscript 2 end subscript end fraction
General
physics-

A bimetallic strip consists of metals X and Y. It is mounted rigidly at the base as shown. The metal X has a higher coefficient of expansion compared to that for metal Y. when bimetallic strip is placed in a cold bath

The metal X has a higher coefficient of expansion compared to that for metal Y so, on placing bimetallic strip in a cold bath, X will shrink more than Y. Hence, the strip will bend towards the left.

A bimetallic strip consists of metals X and Y. It is mounted rigidly at the base as shown. The metal X has a higher coefficient of expansion compared to that for metal Y. when bimetallic strip is placed in a cold bath

physics-General
The metal X has a higher coefficient of expansion compared to that for metal Y so, on placing bimetallic strip in a cold bath, X will shrink more than Y. Hence, the strip will bend towards the left.
General
physics-

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T subscript 2 end subscript and T subscript 1 end subscript left parenthesis T subscript 2 end subscript greater than T subscript 1 end subscript right parenthesis. The rate of heat transfer through the slab, in a steady state is open parentheses fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close parentheses f, with f equals to

Let the temperature of common interface be T ℃. Rate of heat flow
H equals fraction numerator Q over denominator t end fraction equals blank fraction numerator K A increment T over denominator l end fraction
therefore H subscript 1 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 1 end subscript=fraction numerator 2 K A left parenthesis T minus T subscript 1 right parenthesis end subscript over denominator 4 x end fraction
And H subscript 2 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 2 end subscript=fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
In steady state, the rate of heat flow should be same in whole system i e comma
H subscript 1 end subscript equals blank H subscript 2 end subscript
rightwards double arrow fraction numerator 2 K A left parenthesis T minus T subscript 1 end subscript right parenthesis over denominator 4 x end fraction = fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
rightwards double arrow fraction numerator T minus T subscript 1 end subscript over denominator 2 end fraction equals blank T subscript 2 end subscript minus T
rightwards double arrow T minus T subscript 1 end subscript equals blank 2 T subscript 2 end subscript minus 2 T
rightwards double arrow T equals fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction (i)
Hence, heat flow from composite slab is
H equals blank fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
= fraction numerator K A over denominator x end fraction open parentheses T subscript 2 end subscript minus fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction close parentheses equals blank fraction numerator K A over denominator 3 x end fraction blank left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis (ii)
Accordingly, H equals open square brackets fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close square brackets f (iii)
By comparing Eqs. (ii) and (iii), we get
rightwards double arrow f equals blank fraction numerator 1 over denominator 3 end fraction

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T subscript 2 end subscript and T subscript 1 end subscript left parenthesis T subscript 2 end subscript greater than T subscript 1 end subscript right parenthesis. The rate of heat transfer through the slab, in a steady state is open parentheses fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close parentheses f, with f equals to

physics-General
Let the temperature of common interface be T ℃. Rate of heat flow
H equals fraction numerator Q over denominator t end fraction equals blank fraction numerator K A increment T over denominator l end fraction
therefore H subscript 1 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 1 end subscript=fraction numerator 2 K A left parenthesis T minus T subscript 1 right parenthesis end subscript over denominator 4 x end fraction
And H subscript 2 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 2 end subscript=fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
In steady state, the rate of heat flow should be same in whole system i e comma
H subscript 1 end subscript equals blank H subscript 2 end subscript
rightwards double arrow fraction numerator 2 K A left parenthesis T minus T subscript 1 end subscript right parenthesis over denominator 4 x end fraction = fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
rightwards double arrow fraction numerator T minus T subscript 1 end subscript over denominator 2 end fraction equals blank T subscript 2 end subscript minus T
rightwards double arrow T minus T subscript 1 end subscript equals blank 2 T subscript 2 end subscript minus 2 T
rightwards double arrow T equals fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction (i)
Hence, heat flow from composite slab is
H equals blank fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
= fraction numerator K A over denominator x end fraction open parentheses T subscript 2 end subscript minus fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction close parentheses equals blank fraction numerator K A over denominator 3 x end fraction blank left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis (ii)
Accordingly, H equals open square brackets fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close square brackets f (iii)
By comparing Eqs. (ii) and (iii), we get
rightwards double arrow f equals blank fraction numerator 1 over denominator 3 end fraction
General
physics-

A solid material is supplied with heat at constant rate and the temperature of the material changes as shown. From the graph, the false conclusion drawn is

A solid material is supplied with heat at constant rate and the temperature of the material changes as shown. From the graph, the false conclusion drawn is

physics-General
General
physics-

The energy distribution E with the wavelength left parenthesis lambda right parenthesis for the black body radiation at temperature T blank k e l v i n is shown in the figure. As the temperature is increased the maxima will

According to Wien’s displacement law lambda subscript m end subscript proportional to fraction numerator 1 over denominator T end fraction. Hence, it temperature increases lambda subscript m end subscript decreases i. e. comma peak of the E minus lambda curve shift towards left

The energy distribution E with the wavelength left parenthesis lambda right parenthesis for the black body radiation at temperature T blank k e l v i n is shown in the figure. As the temperature is increased the maxima will

physics-General
According to Wien’s displacement law lambda subscript m end subscript proportional to fraction numerator 1 over denominator T end fraction. Hence, it temperature increases lambda subscript m end subscript decreases i. e. comma peak of the E minus lambda curve shift towards left
General
physics-

A metal rod of length 2 m has cross sectional areas 2 A and A as shown in figure. The ends are maintained at temperatures 100 ℃ and 70 ℃. The temperature at middle point C is

Let theta be temperature of middle point C and in series rate of heat flow is same
rightwards double arrow K open parentheses 2 A close parentheses open parentheses 100 minus theta close parentheses equals K A left parenthesis theta minus 70 right parenthesis
rightwards double arrow 200 minus 2 theta equals theta minus 70 rightwards double arrow 3 theta equals 270 rightwards double arrow theta equals 90 ℃

A metal rod of length 2 m has cross sectional areas 2 A and A as shown in figure. The ends are maintained at temperatures 100 ℃ and 70 ℃. The temperature at middle point C is

physics-General
Let theta be temperature of middle point C and in series rate of heat flow is same
rightwards double arrow K open parentheses 2 A close parentheses open parentheses 100 minus theta close parentheses equals K A left parenthesis theta minus 70 right parenthesis
rightwards double arrow 200 minus 2 theta equals theta minus 70 rightwards double arrow 3 theta equals 270 rightwards double arrow theta equals 90 ℃
General
physics-

The coefficient of thermal conductivity of copper is nine times that of steel. In the composite cylindrical bar show in figure, what will be the temperature at the junction of copper ad steel?

Let the temperature of junction be theta.
open parentheses fraction numerator increment Q over denominator d subscript 1 end subscript end fraction close parentheses subscript c o p p e r end subscript blank equals open parentheses fraction numerator increment Q over denominator increment T end fraction close parentheses subscript s t e e l end subscript
K subscript 1 end subscript A equals blank fraction numerator open parentheses 100 minus theta close parentheses over denominator 18 end fraction equals fraction numerator K subscript 2 end subscript A open parentheses theta minus 0 close parentheses over denominator 6 end fraction
9 K subscript 2 end subscript fraction numerator open parentheses 100 minus theta close parentheses over denominator 3 end fraction equals K subscript 2 end subscript theta
3theta equals 900 minus 9 theta
12theta equals 900
theta equals 75 ℃

The coefficient of thermal conductivity of copper is nine times that of steel. In the composite cylindrical bar show in figure, what will be the temperature at the junction of copper ad steel?

physics-General
Let the temperature of junction be theta.
open parentheses fraction numerator increment Q over denominator d subscript 1 end subscript end fraction close parentheses subscript c o p p e r end subscript blank equals open parentheses fraction numerator increment Q over denominator increment T end fraction close parentheses subscript s t e e l end subscript
K subscript 1 end subscript A equals blank fraction numerator open parentheses 100 minus theta close parentheses over denominator 18 end fraction equals fraction numerator K subscript 2 end subscript A open parentheses theta minus 0 close parentheses over denominator 6 end fraction
9 K subscript 2 end subscript fraction numerator open parentheses 100 minus theta close parentheses over denominator 3 end fraction equals K subscript 2 end subscript theta
3theta equals 900 minus 9 theta
12theta equals 900
theta equals 75 ℃
General
maths-

Which function is shown in graph?

Which function is shown in graph?

maths-General
General
maths-

Which function is shown in graph?

Which function is shown in graph?

maths-General
General
maths-

Which function is shown in graph?

Which function is shown in graph?

maths-General
General
physics-

Five rods of same dimensions are arranged as shown in figure. They have thermal conductivities K subscript 1 end subscript comma blank K subscript 2 end subscript comma blank K subscript 3 end subscript comma blank K subscript 4 end subscript a n d K subscript 5 end subscript. When points A blank a n d blank B are maintained at different temperature, no heat would flow through central rod, if

The equivalent electrical circuit, figure in these cases is of Wheatstone bridge. No current would flow through central rod C D when the bridge is balanced. The condition for balanced Wheatstone bridge is fraction numerator P over denominator Q end fraction equals fraction numerator R over denominator S end fraction (in terms of resistances)
fraction numerator 1 divided by K subscript 1 end subscript over denominator 1. K subscript 2 end subscript end fraction equals fraction numerator 1 divided by K subscript 3 end subscript over denominator 1 divided by K subscript 4 end subscript end fraction or fraction numerator K subscript 2 end subscript over denominator K subscript 1 end subscript end fraction equals fraction numerator K subscript 4 end subscript over denominator K subscript 3 end subscript end fraction
Or K subscript 1 end subscript K subscript 4 end subscript equals K subscript 2 end subscript K subscript 3 end subscript

Five rods of same dimensions are arranged as shown in figure. They have thermal conductivities K subscript 1 end subscript comma blank K subscript 2 end subscript comma blank K subscript 3 end subscript comma blank K subscript 4 end subscript a n d K subscript 5 end subscript. When points A blank a n d blank B are maintained at different temperature, no heat would flow through central rod, if

physics-General
The equivalent electrical circuit, figure in these cases is of Wheatstone bridge. No current would flow through central rod C D when the bridge is balanced. The condition for balanced Wheatstone bridge is fraction numerator P over denominator Q end fraction equals fraction numerator R over denominator S end fraction (in terms of resistances)
fraction numerator 1 divided by K subscript 1 end subscript over denominator 1. K subscript 2 end subscript end fraction equals fraction numerator 1 divided by K subscript 3 end subscript over denominator 1 divided by K subscript 4 end subscript end fraction or fraction numerator K subscript 2 end subscript over denominator K subscript 1 end subscript end fraction equals fraction numerator K subscript 4 end subscript over denominator K subscript 3 end subscript end fraction
Or K subscript 1 end subscript K subscript 4 end subscript equals K subscript 2 end subscript K subscript 3 end subscript