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Units and Measurements physics ncert solutions class 11th Physics 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

3 months ago

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 go

...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 $\sqrt[]{\frac{h G}{c^{5}}}$

less

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= kcxhyGz [ML0T0]= [LT-1]x [ML2T-1]y [M-1L3T-2]z= [My-zLx+2y+3zT-x-y-2z]y-z=1x+2y+3z=0-x-y-2z=0On solving these equation we got x= ½ y= ½ and z= -½So formula will coming out from this is m=k <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>c</mi> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mi>G</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> (ii) L=kcxhyGzao [M0LT0]= [LT-1]x [ML2T-1]y [M-1L3T-2]z= [My-zLx+2y+3zT-x-y-2z]y-z=0x+2y+3z=1-x-y-2z=0After solving we get x= -3/2 y=1/2  and z=1/2We got the formula is L=k <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>h</mi> <mi>G</mi> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> (iii)T= kcxhyGz [M0L0T]= [LT-1]x [ML2T-1]y [M-1L3T-2]z= [My-zLx+2y+3zT-x-y-2z]On comparing powers we got x= -5/2 y=1/2 and z= ½So formula for this would be T= k <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>h</mi> <mi>G</mi> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math>

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The one division of main scale of Vernier calipers reads 1mm and 10 divisions of Vernier scale is equal to the 9 divisions on main scale, When the two jaws of the instrument touch each other, the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies in between 4.1 cm and 4.2cm and 6^th Vernier division coincides with a main scale division. The diameter of the bob will be ________× 10^-2 cm.

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$Δ Q = m L \Rightarrow [L] = \frac{M L^{2} T^{- 2}}{M} = L^{2} T^{- 2}$

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