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Units and Measurements physics ncert solutions class 11th Physics 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

3 months ago

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 = $\frac{π}{3600 \times 180} r a d$

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

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

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

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

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

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 1sec1parsec = ( <math> <mfrac> <mrow> <mrow> <mn>1</mn> <mi>A</mi> <mi>U</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mi>a</mi> <mi>r</mi> <mi>c</mi> <mi> </mi> <mi>s</mi> <mi>e</mi> <mi>c</mi> </mrow> </mrow> </mfrac> </math> )1 deg = 3600 arc sec1 parsec = <math> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>3600</mn> <mo>×</mo> <mn>180</mn> </mrow> </mrow> </mfrac> <mi>r</mi> <mi>a</mi> <mi>d</mi> </math> 1parsec = <math> <mfrac> <mrow> <mrow> <mn>3600</mn> <mo>×</mo> <mn>180</mn> </mrow> </mrow> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> </mfrac> <mi>A</mi> <mi>U</mi> </math> = 2 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> <mi>A</mi> <mi>U</mi> </math> (b) sun’s diameter is 1/20so at 1 parsec star is <math> <mfrac> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>  degree in diameter= 15 <math> <mo>×</mo> </math> 10-5 arc/min with magnification it looks like 15 <math> <mo>×</mo> </math> 10-3(c) <math> <mi> </mi> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>D</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> <mi>a</mi> <mi>r</mi> <mi>s</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>D</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> <mi>a</mi> <mi>r</mi> <mi>t</mi> <mi>h</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> we know that <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>D</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> <mi>a</mi> <mi>r</mi> <mi>t</mi> <mi>h</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>D</mi> </mrow> </mrow> <mrow> <mrow> <mi>s</mi> <mi>u</mi> <mi>n</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>100</mn> </mrow> </mrow> </mfrac> </math> <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>D</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> <mi>a</mi> <mi>r</mi> <mi>s</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>D</mi> </mrow> </mrow> <mrow> <mrow> <mi>s</mi> <mi>u</mi> <mi>n</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>×</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>100</mn> </mrow> </mrow> </mfrac> </math> At 1AU sun’s diameter =1/20Mars diameter = <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>×</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>200</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>400</mn> </mrow> </mrow> </mfrac> </math> AT 1/2AU Mars diameter will 1/2000 but with hundred magnification it is 1/20

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