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5 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Two

Einstein's mass - energy relation emerging out of his famous theory of relativity relates mass (m ) to energy (E ) as E = mc², where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV, where 1 MeV= 1.6*10^-13 J; the masses are measured in unified atomic mass unit (u) where 1u = 1.67 * 10^-27 kg.

(a) Show that the energy equivalent of 1 u is 931.5 MeV.

(b) A student writes the relation as 1 u = 931.5 MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.

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Payal Gupta

Contributor-Level 10

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

(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

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5 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Two

(a) How many astronomical units (A.U.) make 1 parsec?

(b) Consider a sunlike star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears ^too be (1/2)° from the earth. Due to atmospheric fluctuations, eye can't resolve objects smaller than 1 arc minute.

(c) Mars has approximately half of the earth's diameter. When it is closest to the earth it is at about 1/2 A.U. from the earth. Calculate what size it will appear when seen through the same telescope.

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Payal Gupta

Contributor-Level 10

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 = ( $\frac{1 A U}{1 a r c s e c}$ )

1 deg = 3600 arc sec

1 parsec = $\frac{π}{3600 * 180} r a d$

1parsec = $\frac{3600 * 180}{π} A U$ = 2 $* 10^{5} A U$

(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) $\frac{D_{m a r s}}{D_{e a r t h}} = \frac{1}{2}$

we know that $\frac{D_{e a r t h}}{D_{s u n}} = \frac{1}{100}$

$\frac{D_{m a r s}}{D_{s u n}} = \frac{1}{2} * \frac{1}{100}$

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

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5 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Two

In an experiment to estimate the size of a molecule of oleic acid 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of this solution is diluted to 20 mL by adding alcohol. Now 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule. Read the passage carefully and answer the following questions:

(a)

...more

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Payal Gupta

Contributor-Level 10

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

(a) Oleic acid does not dissolve in water hence, it is dissolved in alcohol.

(b) Lycopodium powder spreads over the entire surface of water when it is sprinkled evenly. When a drop of prepared solution is dropped on water, oleic acid does not dissolve in water. Instead it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls.

(c) In each mL of solution volume of oleic acid = 1/20mL $* \frac{1}{20}$ = 1/400mL

(d) Volume of n drops of this solution of oleic acid can be calculated by burette and measuring cylinder

(e) If n drops of

...more

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

(a) Oleic acid does not dissolve in water hence, it is dissolved in alcohol.

(c) In each mL of solution volume of oleic acid = 1/20mL $* \frac{1}{20}$ = 1/400mL

(d) Volume of n drops of this solution of oleic acid can be calculated by burette and measuring cylinder

(e) If n drops of sodium solution make 1mL, the volume of oleic acid in one drop will be $\frac{1}{400 n}$ mL

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5 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Two

An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler's Third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that

T= $\frac{k}{R} \sqrt[]{\frac{r^{3}}{g}}$ ,

where k is a dimensionless constant and g is acceleration due to gravity.

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Payal Gupta

Contributor-Level 10

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

According to kepler's law T² $\propto$ r³

T $\propto r^{3 / 2}$

T $\propto r^{\frac{3}{2}} R^{a} g^{b}$

T $\propto k r^{\frac{3}{2}} R^{a} g^{b}$

[M⁰L⁰T]=K [L]^3/2 [L]^a [LT^-2]^b

= k [L^a+b+3/2T^-2b]

On comparing the powers of same terms we get

So a+b+3/2=0

So b=-1/2 and a=-1

T=kr^3/2R^-1g^-1/2

T= $\frac{k}{R} \sqrt[]{\frac{r^{3}}{g}}$

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5 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Two

An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s Third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that

T= $\frac{k}{R} \sqrt[]{\frac{r^{3}}{g}}$ ,

where k is a dimensionless constant and g is acceleration due to gravity.

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5 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Two

If velocity of light c, Planck's constant h and gravitational constant G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

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Payal Gupta

Contributor-Level 10

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

(i) dimension of h = [ML²T^-1]

And c= [LT^-1] and dimension of G= [M^-1L³T^-2]

Let m= kc^xh^yG^z

[ML⁰T⁰]= [LT^-1]^x [ML²T^-1]^y [M^-1L³T^-2]^z

= [M^y-zL^x+2y+3zT^-x-y-2z]

y-z=1

x+2y+3z=0

-x-y-2z=0

On solving these equation we got x= ½ y= ½ and z= -½

So formula will coming out from this is m=k $\sqrt[]{\frac{c h}{G}}$

(ii) L=kc^xh^yG^z

ao [M⁰LT⁰]= [LT^-1]^x [ML²T^-1]^y [M^-1L³T^-2]^z

= [M^y-zL^x+2y+3zT^-x-y-2z]

y-z=0

x+2y+3z=1

-x-y-2z=0

After solving we get x= -3/2 y=1/2 and z=1/2

We got the formula is L=k $\sqrt[]{\frac{h G}{c^{3}}}$

(iii)T= kc^xh^yG^z

[M⁰L⁰T]= [LT^-1]^x [ML²T^-1]^y [M^-1L³T^-2]^z

= [M^y-zL^x+2y+3zT^-x-y-2z]

On comparing powers we got x= -5/2 y=1/2 an

...more

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

(i) dimension of h = [ML²T^-1]

And c= [LT^-1] and dimension of G= [M^-1L³T^-2]

Let m= kc^xh^yG^z

[ML⁰T⁰]= [LT^-1]^x [ML²T^-1]^y [M^-1L³T^-2]^z

= [M^y-zL^x+2y+3zT^-x-y-2z]

y-z=1

x+2y+3z=0

-x-y-2z=0

On solving these equation we got x= ½ y= ½ and z= -½

So formula will coming out from this is m=k $\sqrt[]{\frac{c h}{G}}$

(ii) L=kc^xh^yG^z

ao [M⁰LT⁰]= [LT^-1]^x [ML²T^-1]^y [M^-1L³T^-2]^z

= [M^y-zL^x+2y+3zT^-x-y-2z]

y-z=0

x+2y+3z=1

-x-y-2z=0

After solving we get x= -3/2 y=1/2 and z=1/2

We got the formula is L=k $\sqrt[]{\frac{h G}{c^{3}}}$

(iii)T= kc^xh^yG^z

[M⁰L⁰T]= [LT^-1]^x [ML²T^-1]^y [M^-1L³T^-2]^z

= [M^y-zL^x+2y+3zT^-x-y-2z]

On comparing powers we got x= -5/2 y=1/2 and z= ½

So formula for this would be T= k $\sqrt[]{\frac{h G}{c^{5}}}$

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New answer posted

5 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Two

In the expression P = E l²m^-5 G^-2, E, m, l and G denote energy, mass, angular momentum and gravitational constant, respectively. Show that P is a dimensionless quantity.

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Payal Gupta

Contributor-Level 10

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

Dimensions of energy is E= [ML²T^-2]

Mass m= [M]

Dimension of E= [ML²T^-2]

Dimensions of L= [ML^-2T^-1]

Dimensions of G= [M^-1L³T^-2]

By using these values [P]= [ML²T^{-2] $* [M L^{2} T^{- 1}]$ 2 $[{M]}^{- 5} * [M^{- 1} L^{3} T^{- 2}]$ -2}

= [M^1+2-5+2L^2+4-6T^-2-2+4]

= [M⁰L⁰T⁰]

After we know that P is dimensionless quantity

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5 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Two

A physical quantity X is related to four measurable quantities a, b, c and d as follows:

X = a² b³ c^5/2 d^-2

The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4%, respectively. What is the percentage error in quantity X ? If the value of X calculated on the basis of the above relation is 2.763, to what value should you round off the result.

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Payal Gupta

Contributor-Level 10

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

As we know that X= a² b³ c^5/2 d^-2

Maximum percentage error in X is $\frac{? X}{X} * 100 = \pm [2 (\frac{? a}{a} * 100) + 3 (\frac{? b}{b} * 100) + \frac{5}{2} (\frac{? c}{C} * 100) + 2 (\frac{? d}{d} * 100)]$

= $\pm [2 (1) + 3 (2) + \frac{5}{2} (3) + 2 (4)] %$

$\pm [2 + 6 + \frac{15}{2} + 8]$

$= \pm 23.5 %$

Mean absolute error in X= $\pm 0.24$ rounding off to significant value.

And calculated value would be 2.8 rounding off upto two digits.

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New answer posted

5 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Two

The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as

V= $\frac{π}{8} \frac{p r^{4}}{η l}$

where P is the pressure difference between the two ends of the pipe and η is coefficent of viscosity of the liquid having dimensional formula ML^-1 T^-1. Check whether the equation is dimensionally correct.

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Payal Gupta

Contributor-Level 10

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

Rate of flow is equal to V= $\frac{π}{8} \frac{p r^{4}}{η l}$

Dimensions of V or LHS= volume/time=L³/T= [L³T^-1]

Dimensions of P= [ML^-1T^-2]

Dimensions of $η$ = [ML^-1T^-1]

Dimensions of L= [L]

Dimensions of r= [L]

Dimensions of RHS= $\frac{[M L^{- 1} T^{- 2}]}{[M L^{- 1} T^{- 1}]} \frac{[L^{4}]}{[L]} = [L^{3} T^{- 1}]$

So they are in equal in dimensions.

So equation is correct dimensionally.

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New answer posted

5 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Two

A new system of units is proposed in which unit of mass is α kg, unit of length β m and unit of time γ s. How much will 5 J measure in this new system?

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Payal Gupta

Contributor-Level 10

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

Energy E= [ML²T^-2]

Let M₁, L₁and T₁ and M₂, L₂and T₂ are fundamental quantities for two units

M₁=1kg and L₁=1m and T₁=1s

M₂= α kg, L₂= β m and T₂= γ s

And n₁u₁=n₂u₂

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