Question

1- गद्यांशं पठित्वा अधोलिखितानां पंच प्रश्नानाम् उत्तराणि लिखत।

पुरा मथुरायाम् एकः नृशंसः नृपतिः अभवत् । तस्य नाम कंसः आसीत् कंसः श्रीकृष्णस्य मातुल: आसीत् । सः देवकीम् वसुदेवम् कारागारे अक्षिपत् । तत्र कारागारे एवं श्रीकृष्णः उत्पन्नः अभवत् वसुदेवः शिशुम् कृष्णम् करण्डके निधाय यमुनायाः नद्याः पारे गोकुलम् अनयत् । ततः सः नन्दस्य पुत्रीम् च गृहीत्वा मथुरां प्रत्यागच्छत् ।)- मथुराया: नृपः कः आसीत् ? *
1 point
श्रीकृष्णः
कंस:
नन्द:
वसुदेव:
(2)- कस्य पितुः नाम वसुदेवः आसीत् ? *
1 point
श्रीकृष्णस्य
कंसस्य
देवक्य:
कस्यापि न
(3)- श्रीकृष्णः कुत्र उत्पन्न: अभवत् ? *
1 point
गृहे
नद्याम्
उद्याने
कारागारे
(4)- वसुदेव: काम् गृहीत्वा मथुरां प्रत्यागच्छत्? *
1 point
श्रीकृष्णम्
कंसम्
नन्दस्य पुत्रीम्
कामपि न
(5)- 'नृपतिः' इति विशेष्य पदस्य किं विशेषणपदं अत्र प्रयुक्तम् ? *
1 point
मथुरायाम्
नृशंसः:
अभवत्
श्रीकृष्ण:​

Answer 1

Answer:

Given :-

An object 4 cm height is placed at a distance of 18 cm from concave mirror having focal length 12 cm.

To Find :-

What is the position, nature and height of the image.

Formula Used :-

$\clubsuit$ Mirror Formula :

$\mapsto \sf\boxed{\bold{\pink{\dfrac{1}{f} =\: \dfrac{1}{u} + \dfrac{1}{v}}}}$

where,

f = Focal Length

u = Object Distance

v = Image Distance

$\clubsuit$ Magnification Formula :

$\mapsto \sf\boxed{\bold{\pink{\dfrac{h_i}{h_o} =\: \dfrac{- v}{u}}}}$

Solution :-

First, we have to find the image distance :

Given :

Object Distance = - 18 cm

Focal Length = - 12 cm

According to the question by using the formula we get,

$\implies \sf \dfrac{1}{- 12} =\: \dfrac{1}{- 18} + \dfrac{1}{v}$

$\implies \sf - \dfrac{1}{12} - \bigg(\dfrac{1}{- 18}\bigg) =\: \dfrac{1}{v}$

$\implies \sf - \dfrac{1}{12} + \dfrac{1}{18} =\: \dfrac{1}{v}$

$\implies \sf \dfrac{- 3 + 2}{36} =\: \dfrac{1}{v}$

$\implies \sf \dfrac{- 1}{36} =\: \dfrac{1}{v}$

By doing cross multiplication we get,

$\implies \sf - v =\: 36(1)$

$\implies \sf - v =\: 36$

$\implies \sf \bold{\purple{v =\: - 36\: cm}}$

Now, we have to find the height of image :

Given :

Height of Object = 4 cm

Image Distance = - 36

Object Distance = 18

According to the question by using the formula we get,

$\longrightarrow \sf \dfrac{h_i}{4} =\: \dfrac{- (- 36)}{18}$

$\longrightarrow \sf \dfrac{h_i}{4} =\: \dfrac{36}{18}$

By doing cross multiplication we get,

$\longrightarrow \sf h_i \times 18 =\: 4 \times 36$

$\longrightarrow \sf 18h_i =\: 144$

$\longrightarrow \sf h_i =\: \dfrac{\cancel{144}}{\cancel{18}}$

$\longrightarrow \sf h_i =\: \dfrac{8}{1}$

$\longrightarrow \sf\bold{\red{h_i =\: 8\: cm}}$

${\small{\bold{\underline{\therefore\: The\: height\: of\: the\: image\: is\: 8\: cm\: .}}}}$

Hence, the nature of the image is real and inverted.

Answer 2

Answer:

Given :-

An object 4 cm height is placed at a distance of 18 cm from concave mirror having focal length 12 cm.

To Find :-

What is the position, nature and height of the image.

Formula Used :-

$\clubsuit$ Mirror Formula :

$\mapsto \sf\boxed{\bold{\pink{\dfrac{1}{f} =\: \dfrac{1}{u} + \dfrac{1}{v}}}}$

where,

f = Focal Length

u = Object Distance

v = Image Distance

$\clubsuit$ Magnification Formula :

$\mapsto \sf\boxed{\bold{\pink{\dfrac{h_i}{h_o} =\: \dfrac{- v}{u}}}}$

Solution :-

First, we have to find the image distance :

Given :

Object Distance = - 18 cm

Focal Length = - 12 cm

According to the question by using the formula we get,

$\implies \sf \dfrac{1}{- 12} =\: \dfrac{1}{- 18} + \dfrac{1}{v}$

$\implies \sf - \dfrac{1}{12} - \bigg(\dfrac{1}{- 18}\bigg) =\: \dfrac{1}{v}$

$\implies \sf - \dfrac{1}{12} + \dfrac{1}{18} =\: \dfrac{1}{v}$

$\implies \sf \dfrac{- 3 + 2}{36} =\: \dfrac{1}{v}$

$\implies \sf \dfrac{- 1}{36} =\: \dfrac{1}{v}$

By doing cross multiplication we get,

$\implies \sf - v =\: 36(1)$

$\implies \sf - v =\: 36$

$\implies \sf \bold{\purple{v =\: - 36\: cm}}$

Now, we have to find the height of image :

Given :

Height of Object = 4 cm

Image Distance = - 36

Object Distance = 18

According to the question by using the formula we get,

$\longrightarrow \sf \dfrac{h_i}{4} =\: \dfrac{- (- 36)}{18}$

$\longrightarrow \sf \dfrac{h_i}{4} =\: \dfrac{36}{18}$

By doing cross multiplication we get,

$\longrightarrow \sf h_i \times 18 =\: 4 \times 36$

$\longrightarrow \sf 18h_i =\: 144$

$\longrightarrow \sf h_i =\: \dfrac{\cancel{144}}{\cancel{18}}$

$\longrightarrow \sf h_i =\: \dfrac{8}{1}$

$\longrightarrow \sf\bold{\red{h_i =\: 8\: cm}}$

${\small{\bold{\underline{\therefore\: The\: height\: of\: the\: image\: is\: 8\: cm\: .}}}}$

Hence, the nature of the image is real and inverted.

Answer 3

Answer:

Given :-

An object 4 cm height is placed at a distance of 18 cm from concave mirror having focal length 12 cm.

To Find :-

What is the position, nature and height of the image.

Formula Used :-

$\clubsuit$ Mirror Formula :

$\mapsto \sf\boxed{\bold{\pink{\dfrac{1}{f} =\: \dfrac{1}{u} + \dfrac{1}{v}}}}$

where,

f = Focal Length

u = Object Distance

v = Image Distance

$\clubsuit$ Magnification Formula :

$\mapsto \sf\boxed{\bold{\pink{\dfrac{h_i}{h_o} =\: \dfrac{- v}{u}}}}$

Solution :-

First, we have to find the image distance :

Given :

Object Distance = - 18 cm

Focal Length = - 12 cm

According to the question by using the formula we get,

$\implies \sf \dfrac{1}{- 12} =\: \dfrac{1}{- 18} + \dfrac{1}{v}$

$\implies \sf - \dfrac{1}{12} - \bigg(\dfrac{1}{- 18}\bigg) =\: \dfrac{1}{v}$

$\implies \sf - \dfrac{1}{12} + \dfrac{1}{18} =\: \dfrac{1}{v}$

$\implies \sf \dfrac{- 3 + 2}{36} =\: \dfrac{1}{v}$

$\implies \sf \dfrac{- 1}{36} =\: \dfrac{1}{v}$

By doing cross multiplication we get,

$\implies \sf - v =\: 36(1)$

$\implies \sf - v =\: 36$

$\implies \sf \bold{\purple{v =\: - 36\: cm}}$

Now, we have to find the height of image :

Given :

Height of Object = 4 cm

Image Distance = - 36

Object Distance = 18

According to the question by using the formula we get,

$\longrightarrow \sf \dfrac{h_i}{4} =\: \dfrac{- (- 36)}{18}$

$\longrightarrow \sf \dfrac{h_i}{4} =\: \dfrac{36}{18}$

By doing cross multiplication we get,

$\longrightarrow \sf h_i \times 18 =\: 4 \times 36$

$\longrightarrow \sf 18h_i =\: 144$

$\longrightarrow \sf h_i =\: \dfrac{\cancel{144}}{\cancel{18}}$

$\longrightarrow \sf h_i =\: \dfrac{8}{1}$

$\longrightarrow \sf\bold{\red{h_i =\: 8\: cm}}$

${\small{\bold{\underline{\therefore\: The\: height\: of\: the\: image\: is\: 8\: cm\: .}}}}$

Hence, the nature of the image is real and inverted.