NCERT Exemplar Class 11 Physics Chapter 2 Units and Measurements

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(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 = $\frac{\pi }{3600\times 180}rad$

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

(b) sun’s diameter is 1/2⁰

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

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

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

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

At 1AU sun’s diameter =1/2⁰

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

AT 1/2AU

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

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

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

E= 939.4MeV

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

= 1uc²

= 931.5MeV

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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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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] $\times \left[\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{-1}\right]$ 2 $[{\mathrm{M}]}^{-5}\times [{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}]$ -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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(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 $\times \frac{1}{20}$ = 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 $\frac{1}{400n}$ mL

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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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Distance of galaxy = 10²⁵m

Speed of light = 3 $\times$ 10⁸ m/s

T = distance /speed = $\frac{{10}^{25}}{3\times {10}^8}\approx \frac{1}{3}\times {10}^{17}$

$=\frac{10}{3}\times {10}^{16}s$

$3.33\times {10}^{16}$ s

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t₁=39.6s, t₂=39.9s and t₃=39.5s

Least count of measuring instrument =0.1s

Precision in the measurement =least count of the measuring instrument =0.1s

So mean value of 20 observation

Mean value of t= $\frac{t_1+t_2+t_3}{3}=\frac{39.6+39.9+39.5}{3}=39.7s$

Absolute error in

 t₁= t-t₁ = 39.7-39.6 =0.1s

In t₂= t-t₂ = 39.7-39.9=-0.2s

In t₃= t-t₁ = 39.7-39.5= 0.2s

Mean error = $\frac{0.1+0.2+0.2}{3}=0.17s\approx 0.2$

Accuracy in the measurement = $?0.2s$

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(b) The sum of these numbers is 663.821. The number with least decimal places is 227.2 is correct to only one decimal places and the closest number to it 664.

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Energy E= [ML²T^-2]

Let M₁, L₁ and T₁ and M₂, L₂ and 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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Rate of flow is equal to V= $\frac{\pi }{8}\frac{p{r}^4}{\eta l}$

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

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

Dimensions of  $\eta$ = [ML^-1T^-1]

Dimensions of L= [L]

Dimensions of r= [L]

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

So they are in equal in dimensions.

So equation is correct dimensionally.

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As we know that X= a² b³ c^5/2 d^-2

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

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

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

$=\pm 23.5\%$

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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(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 $\sqrt[]{\frac{ch}{G}}$

(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 $\sqrt[]{\frac{hG}{c^3}}$

(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 $\sqrt[]{\frac{hG}{c^5}}$

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

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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 $\times$ 10²³

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

1 amu= 1/12 of mass of carbon atom

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

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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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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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(a)

 l=R_E; r= 60R_E

$\theta =\frac{R_E}{60R_E}=\frac{1}{60}rad=\frac{1}{60}\times \frac{180}{\pi }degree=\frac{3}{\pi }\approx 1^o$

Hence angle subtend is $2^o$

(b) But in 2^nd case it is given that moon is (½)° and earth is 2^o so it become 4 times.

$\frac{diameterofearth}{diameterofmoon}=\frac{2/\pi }{1/2\pi }=4$

(c) To measure parallax sun is about 400 times the earth moon distance hence $\frac{r_{sun}}{r_{moon}}=400$

To finding ratio of

$\frac{D_{sun}}{r_{sun}}=\mathrm{}\frac{D_{sun}}{r_{moon}}$

$\frac{D_{sun}}{D_{moon}}=400$

$\frac{D_{earth}}{D_{moon}}=4$

so $\frac{D_{sun}}{D_{earth}}=100$

so $\frac{D_{sun}}{D_{earth}}=100$

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(a) A wall clock can measure time correctly upto one second .

(b) A stop watch can measure time correctly upto a fraction of a second .

(c) A digital watch upto the same precision

(d) An atomic clock measure more accurately.

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It is given that 50 VSD $\approx 49\mathrm{M}\mathrm{S}\mathrm{D}$

1MSD = 50/49VSD

1VSD= 49/50MSD

Minimum inaccuracy = 1MSD-1VSD= 1MSD-49/50MSD=1/50 MSD

1MSD= 0.5mm

Minimum inaccuracy = 1/50 $\times$ 0.5mm=0.01mm

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R_me= distance of moon from earth

R_se=distance of sun from earth

Let angle made by sun and moon is .

$\theta =\frac{A_{sun}}{R_{se}^2}=\frac{A_{moon}}{R_{me}^2}$

$\theta =\frac{\pi R_s^2}{R_{se}^2}=\frac{\pi R_m^2}{R_{me}^2}$

( $\frac{R_s}{R_{se}}$ )2 = ( $\frac{R_m}{R_{me}}$ )2

( $\frac{R_s}{R_m}$ )= ( $\frac{R_{se}}{R_{me}}$ )

Hence radius of sun and moon represents their sizes respectively.

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F= [MLT^-2]=100N…….1

Length L= 10m………….2

Time t= 100s……….3

Substituting the value of L and T  in eqn 1,2,3

M $\times 10\times {100}^{-2}=100$

$\frac{M\times 10}{100\times 100}=100$

M=100 $\times 1000kg$

M= 10⁵kg

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(a) Plane angle = l/r radian its si unit is radian but no dimension formula.

(b) Strain= $\frac{?L}{L}=\frac{changeinlength}{length}$ so no dimension

And g =9.8m/s²

(c) Gravitational constant G= 6.67 $\times {10}^{-11}N{m}^2/{kg}^2$

(d) Reynold’s number is a number having no dimension.

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As we know, $\theta =\frac{l}{r}radian$

$\theta =\frac{\pi }{6}=\frac{l}{31}cm$

L=31 $\times \frac{\pi }{6}cm=\frac{31\times 3.14}{6}cm=16.22cm$

Rounding off to three significant figures it would be 16.2cm

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The formula for solid angle is

 solid angle = area/ (distance)²

= $\frac{1{cm}^2}{5{cm}^2}=\frac{1}{25}=4\times {10}^{-2}steridian$

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For dimensional consistency LHS = RHS

[L]= [A]=L

As wt-kx should be dimensionless [wt]= [kx]=1

w [T]=k [L]=1

So w= [T]^-1 and k= [L]^-1

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(b) The zero after points are not significant. The number can be written as 0.069, Hence number of significant figure are four.

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(c) As we know that formula for density is

density = $\frac{mass}{volume}=\frac{4.237g}{2.5{cm}^3}=1.6948$  so density is 1.7g/cm³

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(d) Rounding off 2.745 to 3 significant figures would be 2.74 so result after in three significant figures is 2.74.

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(a) Given length = (1.62 $\pm 0.1$ )cm

And breadth b= (10.1 $\pm 0.1$ )cm

Area = l $\times$ b=163.62 cm²

But when we rounding off to three significant figures then it would be =164cm²

$\frac{?A}{A}=\frac{?l}{l}+\frac{?b}{b}=\frac{0.1}{16.2}+\frac{0.1}{10.1}=\frac{2.63}{163.62}$

$so?A=A\times \frac{2.63}{163.62}=16.62\times \frac{2.63}{163.32}$ = 2.63cm²

So by rounding it off it become 3cm²

Area A= (164 $\pm 3$ )cm²

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(c) (a) work =force $\times distance=$ [MLT^-2] [L]= [ML²T^-2]

Torque= force (distance)= [ML²T^-2]

(b) angular momentum = mvr = [MLT^-1] [L]= [ML²T-1]

Planck’s constant= E/V= [ML²T^-2]/ [T^-1]= [ML²T^-1]

(c) tension=force= [MLT^-2]

Surface tension = force/length= [MLT^-2]/L= [ML⁰T^-2]

(d) impulse =force (time)= [MLT^-2] [T]= [MLT^-1]

linear momentum = mass (velocity)= [M] [LT^-1]

They both have different formula.

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(a) As it is given that

A=2.5ms^{-1 $\pm$} 0.5ms^-1,

B= 0.10s $\pm$ 0.01s

X= AB= (2.5) (0.10)=0.25m

$\frac{?x}{x}=\frac{0.5}{2.5}+\frac{0.01}{0.10}$

$=\frac{0.05+0.025}{0.25}=\frac{0.075}{0.25}$

After rounding off two significant figures

AB= (0.25 $\pm 0.08$ )m

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(d) According to question

A= 1.0m $\pm$ 0.2m, B= 2 $\pm 0.2$ m

Y= $\sqrt[]{AB}=\sqrt[]{1.0\left(2.0\right)}=1.414m$

After rounding off 1.4m

= $\frac{?Y}{Y}=\frac{1}{2}\left. [\frac{?A}{A}+\frac{?B}{B}\right]=\frac{1}{2}\left. [\frac{0.2}{1.0}+\frac{0.2}{2.0}\right]=\frac{0.6}{2\times 2.0}$

 = $\frac{0.6Y}{2\left(2.0\right)}=0.212$

 After rounding off it becomes 0.2m

Thus the correct value of $\sqrt[]{AB}$ =r+dr

 so its value become 1.4 $\pm$ 0.2m

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(a) All the given value are correct upto two decimal places. Here 5.00mm has the smallest unit and the error in 5.00mm is least so the smallest value has less error so 5.00mm is the most precise value.

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(a) l = 5cm

So by checking it

$?$ l₁ = 5-4.9 = 0.1cm

$?$ l₂= 5-4.805 = 0.195cm

$?$ l₃= 5.25-5 = 0.25cm

$?$ l₄= 5.4-5 = 0.4cm

now $?l$ = 5-4.9= 01cm

has the least error hence 4.9 is the most precise value.

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(c) Y= 1.9×10¹¹N/m²

1N=10⁵dyne

Y= 1.9×10¹¹×10⁵dyne/100²cm²

1m= 100 cm

Y= 1.9 $\times {10}^{11}\times {10}^{5}dyne/{100}^{2}cm$

Y= 1.9 $\times {10}^{16-4}dyne/{cm}^2$

Y= 1.9 $\times$ 10¹²dyne/cm²

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(d) E=  $\propto {\rho }^a{A}^b{T}^c$

E= $=k{\rho }^a{A}^b{T}^c$

E= [ML²T^-2] and $\rho =\left. [ML{T}^{-1}\right]$

A= [L²] and T= [T]

E= [k] [ $\rho$ ]^a [A]^b [T]^c

ML²T^-2= [MLT^-1]^a [L²]^b [T]^c

= M^aL^2b+aT^-a+c

by principle of homogeneity we get,

ML²T^-2= M^aL^2b+aT^-a+c

After solving

A=1, 2b+a=2

So 2b+1=2

2b=1, b= ½

C=-2+a, c=-2+1, c=-1

So formula become

E= $\rho$ A^1/2T^-1

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(b, c) for (c) LHS = [L]

RHS= [L/T]=LT^-1

LHS $\ne$ RHS

Hence is c option is not correct.

In option (b) dimension of angle is vt i.e =L

RHS= L.L=L² and LHS = L

LHS $\ne$ RHS

So option b is also not correct

LHS and RHS in both equation B and C are not equal.

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(a, e) It is given that P, Q, R are having different dimensions. Hence cannot be added or subtracted, so we can say that (a) and (e) are not meaningful. We cannot say about the dimensions of product of these quantities. Hence b, c and d may be meaningful.

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(b, d) as we know E= hv

So h=E/v= $\frac{ML{T}^{-2}}{T^{-1}}$ =ML²T^-1

Dimension of linear impulse = dimensions of momentum= [MLT^-1]

As we know that J= $?p$

Angular impulse =tdt= $?L$ = change in angular momentum

Hence angular momentum is [ML²T^-1] same as the dimension of h.

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(a,b,d) h=[ML²T^-1]

[c]=[LT^-1], [m_e]=M

[G]= [M^-1L³T^-2]

[e]=[AT],[m_p]=[M]

[ $\frac{hc}{G}$ ] = $\frac{\left[M{L}^2{T}^{-1}\right]\left[L{T}^{-1}\right]}{M^{-1}{L}^3{T}^{-2}}=M^2$

M= $\sqrt[]{\frac{hc}{G}}$

$\frac{h}{c}=\frac{M{L}^2{T}^{-1}}{L{T}^{-1}}=\left[ML\right]$

L= $\frac{h}{cM}=\frac{h}{c}\sqrt[]{\frac{G}{hc}}=\frac{\sqrt[]{Gh}}{c^{3/2}}$

C= LT^-1

T= $\frac{\left[L\right]}{\left[C\right]}$ = $\frac{\sqrt[]{Gh}}{c^{3/2}c}=\frac{\sqrt[]{Gh}}{c^{5/2}}$

Hence a,b and d any can be used to express L,M and T in terms of three chosen fundamental quantities.

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(a, b) we know that pressure = force/area

Pressure= $\frac{force\times distance}{area\times distance}=\frac{work}{volume}=\frac{energy}{volume}$

So we can say that pressure is also equal to energy per unit volume.

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(b, d) We know that one light year = 9.46 $\times {10}^{11}m$ = distance that light travels in 1 year with a speed 3 $\times {10}^8$ m/s

1 par sec =3.08 $\times {10}^{16}m$ = distance at which average radius of earth’s orbit subtends an angle of par second.

Here second and year represent time.

Slope = $\frac{\Delta L/W}{L}=\frac{\Delta L/L}{W}=\frac{1}{YA}$ $\frac{}{}\frac{}{}\frac{}{}$

$Y=\frac{1}{\left(Slope\right)A}$

$Y=\frac{1}{\left(2\times 10^{-6}\right)\left(0.25\times 10^{-5}\right)}$

$Y=2\times 10^{11}N/m^2$

Independent of are a is case of uniform wire

$\delta =A\left({\mu }_y-1\right)-A^1\left({\mu }_y^1-1\right)$

$=6\left(1.5-1\right)-5\left(1.55-1\right)$

$=\frac{1}{4}$

$?=\overrightarrow{B}.\overrightarrow{A}$

$=\left(3t^3\widehat{j}+3t^2\widehat{k}\right).\left(\pi {\left(1\right)}^2\widehat{k}\right)$

$?=3t^2\pi$

$E_{ind}=\left|\frac{d?}{dt}\right|=6t\pi$

at t = 2, E_ind = 12

$q=C{V}_{100\Omega }$

$=\left(1.1\times 10^{-6}\right)\left(\frac{10}{R+r}R\right)$

$=1.1\times 10^{-6}\left(\frac{10}{110}\times 100\right)$

$=10\mu C$

$C_{eff}=\left[\frac{{\epsilon }_0\left(7\times 4\right)}{4/10}+\frac{5{\epsilon }_0\left(1\times 4\right)}{4/10}\right]\times 10^{-2}$

$C_{eff}=1.2{\epsilon }_0$

Energy = $\frac{1}{2}{C}_{eff}{V}^2$ $\frac{}{}{}_{}{}^{}$

$\frac{1}{2}\left(1.2{\epsilon }_0\right)\left(20\right)\left(20\right)=240{\epsilon }_0$

$\frac{nv}{0.6}=400\&\frac{\left(n+1\right)v}{0.6}=450$

$\Rightarrow \left[\frac{0.6\times 450}{v}+1\right]$ $\frac{v}{0.6}=450$

v = 30

$\Rightarrow \sqrt[]{\frac{T}{\mu }}=30$

^{$\Rightarrow \frac{2700}{\mu }=900$}

^{$\mu =3$}

$P_2-P_{06}=\frac{4T}{6}\&P_1-P_2=\frac{4T}{3}$

$P_1-P_0=\frac{4T}{2}=\frac{4T}{r}$

r = 2cm

.From conservation of energy

mgh = $\frac{1}{2}m{v}^2+\frac{1}{2}l{\omega }^2$ $\frac{}{}{}^{}\frac{}{}{}^{}$

$\Rightarrow mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{m{R}^2}{2}$

10 h = $\frac{16}{2}+\frac{16}{4}$ $\frac{}{}\frac{}{}$

h = 1.2 m = 120 cm

From conservation of Energy

$\frac{1}{2}m{v}_0^2=mgh$

$v_0=10\sqrt[]{2}$

For A to B

$a=-gsin45°=\frac{-10}{\sqrt[]{2}}$

T = t₁ + t₂

$\begin{array}{l}=2\sqrt[]{2}+2\\ =2\left(\sqrt[]{2}+1\right)\end{array}$