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PHYSICS MCQs
Total Questions : 421
Page 1 of 43 pagesThe focal length of the combination is given by
`1/F = 1/f_1 + 1/f_2`
Here : F = 80 cm = 0.8 m., `f_1` = 20 cm = 0.2 m.
`:.` `1/f_2` =` 1/F - 1/f_1`
or `1/f_2 `= `1/0.8 - 1/0.2`
or `1/f_2` = `(1 - 4)/(0.8) = -3/0.8 = - 3.75`
But ` p = 1/f`
So, power `p_2 = 1/f_2`
Hence , `p_2` = - 3.75 D
From the relation , `mu = 1/sin C`
or ` c_v/c_m = 1/sinC`
where C = critical angle = `30^circ`
`c_v` = velocity of light in vacuum = 3 x `10^8` m/s
`c_m` = velocity of light in medium
Therefore , `(3 xx 10^8)/(c_m) = (1)/(sin30^circ)`
or `c_m= 1/2 xx 3 xx 10^8`
= 1.5 x` 10^8` m/s
For small angled prism - `delta m = (mu - 1)A`
where `delta m ` = deviation of light in prism,
`mu` = refractive index = 1.5
A = angle of prism = `6^circ`
Therefore , `delta m` = (1.5 - 1) x `6^circ`
= 0.5 x `6^circ = 3^circ`
Number of images = `(360^circ)/(theta)-1`
= `(360^circ)/(45^circ) - 1`
= 8 - 1 = 7
According to the first law of thermodynamics
`Delta Q = Delta W + Delta u`
Given : `Delta Q = 2 xx 10^3 xx 4.2 J`
`Delta W = 500 J`
So, ` Delta u = 2 xx 10^3 xx 4.2 J - 500 J`
` = 8400 - 500 = 7900 J`
The first law of thermodynamics is basically the law of conservation of energy . It states that if a system absorbs heat `Delta Q` and as a result its internal energy changes by `Delta u` and the system does a work `Delta W`
Therefore , according to this law
`Delta Q = Delta u + Delta W`
By the principle of calorimety Heat lost = heat gain
`Ms Delta theta `= mL
where : m = mass of ice
M = mass of water = 80g
s = specific heat of water = 1 cal/`g^circ C`
`Delta theta` = temperature of water = `30^circ`
L = latent heat of ice = 80 cal/g
So, 80 x 1 x`30^circ` = m x 80
Hence , m = 30 g.
Bernoulli's principle states that if speed of liquid at a place is maximum then at hat place pressure will be minimum Therefore at the extreme narrow portion of the pipe of non -uniform cross section, the speed is maximum , so pressure will be minimum .
We know that E `prop` T
or ` E_1/E_2 = T_1/T_2`
Here `E_1 = E`
`E_2 = 2E`
`T_1 = 20 + 273 = 293 K`
`:.` `(E)/(2E) = 293/_2`
or `T_2 = 586 K`
Hence , `T_2 = 596 - 273 = 313^circ C`
From the relation , `v_(rms)` =` sqrt((3R)/(m))`
or `(v_(rms)He)/(v_(rms) H_2)` = `sqrt(T_1/m_1 xx m_2/T_2)`
or `5/7 = sqrt((T_1 xx 2)/(4 xx (0 + 273)))`
or `(5/7)^2 = (T_1)/(2(273))`
or `T_1 = (546 xx 25)/(49)`
= 278.6 K
Hence , `T_1 = 5.6^circ C`
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