The angle of $1\xb0$ (degree) will be equal to:

(Use $360\xb0=2\mathrm{\pi}\mathrm{rad},1\xb0=60\text{'}\mathrm{and}1\text{'}=60\text{'}\text{'}$)

$\left(1\right)1.034\times {10}^{-3}\mathrm{rad}\phantom{\rule{0ex}{0ex}}\left(2\right)1.745\times {10}^{-2}\mathrm{rad}\phantom{\rule{0ex}{0ex}}\left(3\right)1.524\times {10}^{-2}\mathrm{rad}\phantom{\rule{0ex}{0ex}}\left(4\right)1.745\times {10}^{+3}\mathrm{rad}$

67%

Subtopic: Measurement & Measuring Devices |

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A man wishes to estimate the distance of a nearby tower from him. He stands at point A in front of tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC up to B, a distance of 100 m, and looks at O and C again. Since O is very distant, the direction BO is practically the same as AO; but he finds the line of sight of C shifted from the original line of sight by an angle θ = $40\xb0$ (θ is known as ‘parallax’), the distance of the tower C from his original position A is:

(1) 119 m

(2) 126 m

(3) 320 m

(4) 219 m

From NCERT

Subtopic: Measurement & Measuring Devices |

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The moon is observed from two diametrically opposite points A and B on Earth. The angle θ subtended at the moon by the two directions of observation is $1\xb0$ 54′. Given the diameter of the Earth to be about 1.276 × ${10}^{7}$ m, the distance of the moon from the Earth is:

$\left(1\right)2.91\times {10}^{8}\mathrm{m}\phantom{\rule{0ex}{0ex}}\left(2\right)3.11\times {10}^{9}\mathrm{m}\phantom{\rule{0ex}{0ex}}\left(3\right)3.84\times {10}^{8}\mathrm{m}\phantom{\rule{0ex}{0ex}}\left(4\right)1.91\times {10}^{7}\mathrm{m}$

Subtopic: Measurement & Measuring Devices |

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The Sun’s angular diameter is measured to be 1920′′. The distance D of the Sun from the Earth is $1.496\times {10}^{11}\mathrm{m}.$ The diameter of the Sun:

1. $2.39\times {10}^{9}\mathrm{m}$

2. $0.39\times {10}^{9}\mathrm{m}$

3. $1.39\times {10}^{9}\mathrm{m}$

4. $1.39\times {10}^{10}\mathrm{m}$

53%

Subtopic: Measurement & Measuring Devices |

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If the size of a nucleus (in the range of ${10}^{-15}\mathrm{to}{10}^{-14}\mathrm{m}$) is scaled up to the tip of a sharp pin, what roughly is the size of an atom? Assume tip of the pin to be in the range ${10}^{-5}\mathrm{m}\mathrm{to}{10}^{-4}\mathrm{m}.$

1. $1\mathrm{m}$

2. $10\mathrm{m}$

3. ${10}^{-10}\mathrm{m}$

4. ${10}^{-5}\mathrm{m}$

From NCERT

Subtopic: Measurement & Measuring Devices |

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We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s. The average absolute error and percentage error, respectively, are:

1. 0.22 s and 4

2. 0.11 s and 4

3. 4 and 0.11 s

4. 5 and 0.22 s

Subtopic: Measurement & Measuring Devices |

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