Explore topic-wise InterviewSolutions in Current Affairs.

Trans – continental railways runs across the continent and links its two ends. These were constructed for economic and political reasons to facilitate long runs in different directions. 

The major Trans – Continental railways are:

1. Trans-Siberian Railways. 

2. Canadian Pacific Railways. 

3. Northern United States Intercontinental Railways. 

4. Southern United States Intercontinental Railways. 

5. Australian Intercontinental Railways.

The highest railways density is found in Belgium (1 km per 6.5 sq. lcm area), a European country.

(c) United States of America

The Trans – Siberian railways passes through the cities of St. Petersburg, Moscow, Kazan, Yakaterin bury, Tyumin, Omsk, Novosibirsk, Krasnoyarsk, Chita, Angarsk, Khabarovsk and Vladivostok.

Cairo would be connected to Capetown through Cape – Cairo Railways.

The Alaska Highway connects the town of Edmonton in Canada to Anchorage city, Alaska.

Object distance = u = 20 cm

a) Focal length =f =-30 cm     (\(\because\)  concave mirror )

Images distance = v = ?

Mirror formula  \(\frac{1}{f}= \frac{1}{v}\,+\,\frac{1}{u}\Rightarrow\,-\frac{1}{30} = \frac{1}{v}\,-\,\frac{1}{20}\)

 \(\frac{1}{v} =\frac{1}{20}\,-\,\frac {1}{30}\,\Rightarrow\, \frac{1}{v}=\frac{3\,-\,2}{60}= \frac{1}{60}\Rightarrow v = 60\,cm. \)

Image distance = 60 cm.

So image is virtual formed in the mirror.

b) Magnification = m = \(-\frac{Image \,distance}{Object\,distance} = \frac{-60}{-20}=3\,cm\)

Here '+' represents image is erect and it is 3 times enlarged .

Correct option is C) at C

Correct Answer is: (a) England 

The cities of New Delhi, Mumbai, Benguluru, Chennai, Kolkata and Hyderabad would be connected under the project of Golden Quadrilateral Corridor Scheme.

Correct Answer is: (c) Four

Correct option is C) – 54 cm

Because these mirrors are fitted on the sides of the vehicle, enabling the driver to see traffic behind him/her to facilitate safe driving.

The object distance in both concave as well as convex mirror is negative.

Correct Answer is: (c) Internet

As the object is moved from infinity towards the pole of a concave mirror, the image formed starts shifting from the focus of the mirror towards infinity.

When the object is at infinity, the image is formed at the focus or in the focal plane.As the object is shifted, the image is formed between the focus and the centre of curvature, then at the centre of curvature, then beyond the centre of curvature, then at infinity and finally the image is formed behind the mirror.

A mirror that has very wide field view is convex.

Some public means of communication are radio, television, cinema, GPS, newspapers, magazines, satellite, conferences, meetings, etc.

1. The image is formed behind the mirror. 

2. It is enlarged, he. magnified. 

3. It is virtual. 

4. It is erect.

If the object is placed at focus of a concave mirror, the image is formed at  infinity.

The communication through satellite has emerged as a new feature in communication technology since 1970. United States of America and former U.S.S.R pioneered space research. Artificial satellites are now successfully deployed in the earth’s orbit to connect even the remote corners of the world with limits of site verification. The satellites have reduced the unit cost and time of communication and the cost incurred in communication for 500 km is now equal to the cost incurred in satellite communication for 5000 km.

Goods like crude – oil, refined petroleum products, natural gas, water and milk are transported through pipelines.

According to the question 

Magnification (m) =\(\frac1n\) (image is of \(\frac1{n^{th}}\) the size of object). 

Focal length = f (positive for convex mirror). 

Object distance = -u. 

Now By formula.

m = \(\frac{f}{f-u}\) 

⇒ \(\frac{1}{n} = \frac{f}{f+u}\)

⇒ f + u = nf 

⇒ u = nf – f 

⇒ u = (n-1)f. 

Therefore the object distance from the mirror is (n-1) f. Hence the option C is correct

Pipelines are the latest means of transportation. Crude oil or petroleum is transported to refineries and refined petroleum products are sent to consumer centres by this means of transportation. In addition, natural gas is also transported by pipelines. A high density of pipelines is found in Europe and middle – east countries. Pipelines are used extensively in transporting liquids and gases such as water, petroleum, and natural gas.

For they have uninterrupted flow, water supply through pipelines is familiar to all. Cooking gas or LPG is supplied through pipelines in many parts of the world. In New Zealand, milk is being supplied through pipelines from dairy farms to processing plants. A dense network of piplines is found between the production regions and the consumption regions of united States of America.

Among these, the most famous is the ‘Big Inch’ pipeline, which supplies oil obtained from the coastal wells of Gulf of Mexico to the north – eastern states. In Europe, western Asia, Russia and India, pipelines are used, to join the oil wells with the refineries and internal markets. The pipline from Turkmenistan, a country in central Asia, has been extended to Iran and some parts of China. After the completion of the proposed Iran – Pakistan – India oil and gas pipeline, it will be the world’s longest pipeline.

As petroleum and natural gas are fluid and flowable so they can easily be sent through pipelines from one place to another place.

Given

Height of image(h_i) = \(\frac{height\,of\,object (h_o)}{3}\) focal length of mirror (f) = 12 cm

m = \(\frac{h_i}{h_o}\,=\,\frac{-v}{u}\,=\,\frac{1}{3}\,\implies\) u = - 3v

Using Mirror formula, \(\implies\,\frac{1}{v}\,+\frac{1}{u}\,=\,\frac{1}{f}\)

\(\implies\,\frac{1}{v}\,+\,\frac{1}{-3v}\,=\,\frac{1}{12}\)

\(\implies\,\frac{2}{3v}\,=\,\frac{1}{12}\)

\(\implies\)v = 8 cm

As v is positive, so image is formed at 8 cm behind the mirror

The radius of curvature of the mirror is double the focal length (R = 2f). The image formed at the centre of curvature of the concave mirror is real, inverted and of the same size.

For the given mirror, the center The object distance is equal to the centre of curvature of the concave mirror.

Major pipelines of the world are: 

1. Big Inch Pipeline: This pipeline supplies oil from the coastal wells of the Gulf of Mexico to the northeast states in the United States.

2. Tap Pipeline: This pipeline connects the oil wells near Persion Gulf to a city called Sudan. Its length is 1600 kilometers.

3. Comacon Pipelines: It is located in the former Soviet Union. Oil from Volga and Ural regions is transported to the countries of Eastern Europe through this pipeline.

4. O.I.L. Pipelines: The O.I.L. pipelines has been constructed From Naharkatiya of Assom to Barauni oil refinery in Bihar. Its length is 1157 km. It was extended upto Kanpur in 1966.

5. H. V J. Pipe lines: An important network of another broad pipeline in West India was constructed in Ankleshwar – Koyali, Mumbai High – ICoyali and Hazira – Vijaipur – Jagdishpur. Recently a 1256 km long pipeline was made from Salaya (Gujarat) to .Mathura (U.P.).

6. Tapi project: The much awaited Turkmenistan-Afganistan-Pakistan-India (Tapi) gas pipeline project was inaugurated on December 3, 2015. in Meri city (Turkmenistan) connected to the historic Silk route. It is likely to be completed by 2019, which will be 1814 km long. It will reach Fazilka (India) via Kandhar (Afghanistan 774 km) and Multan (Pakistan 826 km) from Galkynysh (Turkmenistan) area.

According to the question; 

Focal length (f) = 12cm; 

Object distance (u) = -8cm; 

By mirror formula;

\(\frac1v\,+\,\frac1u\,=\,\frac1f\)

⇒ \(\frac1v\,+\,\frac1{-8}\,=\,\frac1{12}\)

⇒ \(\frac1v\,+\,\frac1{8}\,=\,\frac1{12}\)

\(\frac1v\,=\,\frac{3+2}{24}\)

⇒ \(\frac1v\,=\,\frac{5}{24}\)

⇒ v = \(\frac{5}{24}\) = 4.8 cm.

Since v = 4.8cm which is positive hence image is behind the mirror.

The image is formed at the distance of 4.8 cm behind the mirror and is virtual and erect.

A convex mirror always forms a virtual, erect and diminished image for a real object irrespective of its position.

The river Rhine is usable for boating from its origin Dortmand (Netherland) to its end point in Switzerland (Bessie). The distance between Dortmand to Switzerland is 700 km.

(i) Yes. Donating vital organs can make the life of a living person easier.
(ii) Social welfare, kind-hearted, selfless, friendship.
(iii) Kidneys, heart,liver, etc.

The answer is (b) a convex mirror

• Field of convex mirror is more than any type of mirror. 
• Hence full length size of building can be seen by using a convex mirror

False

A convex always produces a diminished image of the object. It can never produce an image larger than the size of the object.

According to the question;

Focal length (f) = -1.5m; 

Object distance (u) = -40cm = 0.40m; (1m = 100 cm) 

By mirror formula;

\(\frac1v+\frac1u=\frac1f\)

⇒ \(\frac1v+\frac1{-0.40}=\frac1{-1.5}\)

⇒ \(\frac1v=\frac1{0.40}-\frac1{1.5}\)

⇒\(\frac1v=\frac{10}{4}-\frac{10}{15}\)

⇒ \(\frac1v=\frac{5}{2}-\frac{2}{3}\)

⇒ \(\frac1v=\frac{15-4}{6}\)

⇒ \(\frac{1}{v}=\frac{11}{6}\)

⇒ v = 0.5454m = 54.54cm (1m = 100 cm). 

Since v = 54.54cm which is positive hence image is behind the mirror.

The image is formed at the distance of 54.54 cm behind the mirror and is virtual and erect.

(i) This is due to the phenomenon of refraction. Light reflected from the fish travels through the water towards the eye. As it passes from the water into the air, it refracts away from the normal because water is denser than air and thus, the fish appears closer.
(ii) Obeying elders, being careful, understanding reasons for natural phenomena.

Refractive index of air, μair = 1; 

Given; 

Refractive index of water w.r.t air, ^aμ_w = \(\frac43\)

Refractive index of glass w.r.t air, ^aμ_g  = \(\frac32\) 

Refractive index of water w.r.t glass = ^gμ_w = \(\frac{^aμ_w}{^aμ_g{^c}}=\frac{\frac43}{\frac32}\)

⇒ ^gμ_w = \(\frac{4\times2}{3\times3}=\frac89\)

∴ Refractive index of water w.r.t glass is \(\frac89\)

Also; 

Refractive index of water w.r.t glass. = \(\frac{1}{Refractive\,index\,of\,glass\,w.r.t\,water.}\)

⇒ Refractive index of glass w.r.t water = \(\frac{1}{Refractive\,index\,of\,water\,w.r.t\,glass}\)

∴ Refractive index of glass w.r.t water = \(\frac{1}{\frac89}=\frac98\)

Refractive index of glass w.r.t water =\(\frac98\)

As the image is obtained on the screen , it is real . 

so, 

Magnification , m = –3 , 

v = 80 cm 

u = ? 

As m = v/u 

so, –3 = 80/u, 

u = –80/3 cm .

From 

1/f = 1/v - 1/u 

=1/80 + 3/80 

= 4/80 

= 1/20 

1/f = 1/20cm 

so, f = 20 cm . 

The lens is convex and image formed at 80 cm from the lens is real and inverted.

1.  v_e = -25 cm

\(\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}\) = \(\frac{1}{-25}-\frac{1}{6.25}\)

∴ u₀ = 5 cm

Length of the tube, L= |v_o| + |u_e|

∴ v_o = 15 - 5 = 10

\(\frac{1}{u_o}=\frac{1}{v_0}-\frac{1}{f_0}\)

= \(\frac{1}{10}-\frac{1}{2}\) 

u_e = -2.5 cm

2.  ∴ v₀ = 15 – f_e = 15 – 6.25 = 8.75

\(\frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0}\)

= \(\frac{1}{8.75}\) - \(\frac{1}{2}\)

u₀ = -2.59 cm

1. v_e = -25 cm

\(\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}\) = \(\frac{1}{-25}-\frac{1}{6.25}\)

∴ u₀ = 5 cm 

Length of the tube, L= |v₀| + |u_e|

∴ v₀ = 15 – 5 = 10

\(\frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0}\) = \(\frac{1}{10}-\frac{1}{2}\)

u_e = -2.5 cm.

2.  ∴ v₀ = 15 – f_e = 15 – 6.25 = 8.75

\(\frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0}\)= \(\frac{1}{8.75}\) - \(\frac{1}{2}\)

u₀ = -2.59 cm

(c) moves away from the lens with a non-uniform acceleration. 

1. For normal adjustment.

M.P. of telescope = \(\frac{f_0}{f_e}=\frac{144}{6}\) = 24

2. The length of the telescope in normal adjustment

L = f_o + f_e

= 144 + 6 = 150 cm.

1. A 

2. A 

3. A 

4. A 

5. Yes, If we place this lens in a medium of higher refractive index than lens.

(b) Light travel from denser to rarer medium. 

Explanation: In total internal reflection, light travel from denser to rarer medium.

(c) smaller

Explanation:

velocity of light decreases in a medium. Hence λ decrease in a medium (v ∝ λ).

1. For convex lens R₁ = +ve and R₂ = -ve

\(\frac{1}{f}=\Big(\frac{n_2}{n_1}-1\Big)\Big(\frac{1}{R_1}+\frac{1}{R_2}\Big)\)

2.  Yes.

P = \(\frac{1}{f}\) = \(\Big(\frac{n_2}{n_1}-1\Big)\Big(\frac{1}{R_1}+\frac{1}{R_2}\Big)\)

From the above equation it is clear that, Pα \(\frac{n_2}{n_1}\)

lense in air, \(\frac{n_{glass}}{n_{air}}\) = \(\frac{1.5}{1}\) = 1.5

lense in water, \(\frac{n_{glass}}{n_{water}}\) = \(\frac{1.5}{1.33}\) = 1.12

In water,\(\frac{n_2}{n_1}\)  is less. Hence power decreases.

3. P = +ve, converging

P = -ve, diverging

P = 0, Plane glass

1.   \(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\)

\(\frac{1}{f} = \frac{+1}{10}+\frac{-1}{5}\)

= \(\frac{5-10}{50}\) = \(\frac{-1}{10}\)

f = -10 cm

Effective focal length is negative. Hence this lens is diverging.

2. Effective focal length becomes positive Hence the lens will act as converging.

Focal length of convex lens f₁ = 25 cm

Focal length of concave lens f₂ = -25 cm

Power of combination in diopters,

P = P₁ + P₂ = \(\frac{100}{f_1}+\frac{100}{f_2}\) = \(\frac{100}{25}-\frac{100}{25}\) = 0

Answer is (c) 5 c

\(\frac{1}{f_{air}}=\Big(\frac{1.5}{1}-1\Big)\) \(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) ......(1)

\(\frac{1}{f_{liquid}}=\Big(\frac{1.5}{1.25}-1\Big)\)\(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) ......(2)

eq(2)/eq(1)

\(\frac{f_{air}}{f_{liquid}}\) = \(\frac{2}{5}\) , \(f_{liquid}=\frac{5}{2}f_{air}=\frac{5}{2}\times2\) =  5cm

1. m = \(\frac{h_0}{h_1}\)  \(\frac{-1.4}{0.7}\) = -2

2. Real, inverted, magnified 

3. m = \(\frac{f}{f+u}\) ; -2 = \(\frac{f}{f+-30}\) ,f = 20 cm