Physics NCERT Exemplar Solutions Class 11th Chapter Two

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Because they cannot be derived from one another as they are independent quantities.

Also we can say that all other quantities in mechanics can be expressed in terms of length mass and time.

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Since f ( $\theta )$ is a sum of different powers of $\theta$  an we know that it is a dimensionless quantity. By principle of homogeneity as RHS is dimensionless. Hence LHS should also be dimensionless.

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One atomic mass unit is 1/12 of the mass of carbon atom

Mass of one carbon atom = 12g

Number of atoms in one mole= Avogadro's number

= 6.023 $*$ 10²³

Mass of one ₆C¹² atom = $\frac{12}{6.023*{10}^{23}}g$

1 amu= 1/12 of mass of carbon atom

1 amu= ( $\frac{1}{12}*\frac{12}{6.023*{10}^{23}}$ )g= 1.67 $*{10}^{-27}kg$

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A mass spectrograph is used for measuring the mass of atoms and molecules.

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Radius of atom = 1A^o=10^-10m

Radius of nucleus = 10^-15m

Volume of atom=V_A= 4/3 $\pi$ r³_a

Volume of nucleus V_N=4/3 $\pi$ r³_N

To finding the ratio

$\frac{V_A}{V_N}=\frac{\frac{4}{3}\pi r_a^3}{\frac{4}{3}\pi r_N^3}=\frac{r_a}{r_n}={10}^{15}$

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The value of any given physical quantity may very over a wide range. Therefore different units of same physical quantity are required. As we know that to measuring distance of an object we need cm m, km, light year etc.

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(a) 1 amu = 1.67 $*$ 10^-27kg

And energy E=mc²= (1.67 $*$ 10^-27) (3 $*{10}^8$ )²J

E= 939.4MeV

(b) So dimensionally correct relation will be E=amu $*$ c²

= 1uc²

= 931.5MeV

This is a long answer type question as classified in NCERT Exemplar

(a) 1 parsec =distance at which 1AU long arc subtends an angle of 1sec

1parsec = ( $\frac{1AU}{1arcsec}$ )

1 deg = 3600 arc sec

1 parsec = $\frac{\pi }{3600*180}rad$

1parsec = $\frac{3600*180}{\pi }AU$ = 2 $*{10}^{5}AU$

(b) sun's diameter is 1/2⁰

so at 1 parsec star is $\frac{1/2}{2*{10}^5}$  degree in diameter= 15 $*$ 10^-5 arc/min with magnification it looks like 15 $*$ 10^-3

(c) $\frac{D_{mars}}{D_{earth}}=\frac{1}{2}$

we know that $\frac{D_{earth}}{D_{sun}}=\frac{1}{100}$

$\frac{D_{mars}}{D_{sun}}=\frac{1}{2}*\frac{1}{100}$

At 1AU sun's diameter =1/2⁰

Mars diameter = $\frac{1}{2}*\frac{1}{200}=\frac{1}{400}$

AT 1/2AU

 Mars diameter will 1/200⁰ but with hundred magnification it is 1/2⁰