Question

1. Find out the focal length of concave mirror.

A.10 B.15 C.20 D.25

2. Which set of observation is incorrect?

A.3 B.4 C.5 D.6


3.In which set Ram gets the image of candle flame highly diminished?

A.1 B.2 C.3 D.4


4. In which observation he gets the same size of image as candle flame?

A.1 B.2 C.3 D.4





this is case study question if you know please answer​

Answer 1

Answer:

i) The curved surface area of cylinderical petrol tank is 59.4 m².

ii) The steel actually used to make the tank is 95.04 m².

Step-by-step-explanation:

We have given that,

For a cylinderical petrol tank,

Diameter ( d ) = 4.2 m

∴ Radius ( r ) = d ÷ 2 = 4.2 ÷ 2 = 2.1 m

Height ( h ) = 4.5 m

We have to find,

i) Curved surface area of tank

ii) Total steel used to make tank

i)

Now, we know that,

\displaystyle{\pink{\sf\:Curved\:surface\:area\:of\:cylinder\:=\:2\:\pi\:r\:h}}Curvedsurfaceareaofcylinder=2πrh

\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:2\:\times\:\dfrac{22}{\cancel{7}}\:\times\:\cancel{2.1}\:\times\:4.5}⟹CSAcylinder=2×722×2.1×4.5

\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:2\:\times\:22\:\times\:0.3\:\times\:4.5}⟹CSAcylinder=2×22×0.3×4.5

\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:22\:\times\:0.3\:\times\:9}⟹CSAcylinder=22×0.3×9

\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:198\:\times\:0.3}⟹CSAcylinder=198×0.3

\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:59.4\:m^2}⟹CSAcylinder=59.4m2

\therefore\:\underline{\boxed{\red{\sf\:Curved\:surface\:area\:of\:cylinder\:=\:59.4\:m^2\:}}}∴Curvedsurfaceareaofcylinder=59.4m2

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ii)

Now,

The tank is closed with steel.

We have to find how much steel was used to make the tank.

Now,

\displaystyle{\pink{\sf\:Total\:surface\:area\:of\:cylinder\:=\:2\:\pi\:r\:(\:r\:+\:h\:)}}Totalsurfaceareaofcylinder=2πr(r+h)

\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:2\:\times\:\dfrac{22}{\cancel{7}}\:\times\:\cancel{2.1}\:(\:2.1\:+\:4.5\:)}⟹TSAcylinder=2×722×2.1(2.1+4.5)

\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:2\:\times\:22\:\times\:0.3\:\times\:6.6}⟹TSAcylinder=2×22×0.3×6.6

\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:22\:\times\:0.3\:\times\:13.2}⟹TSAcylinder=22×0.3×13.2

\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:22\:\times\:3.96}⟹TSAcylinder=22×3.96

\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:87.12\:m^2}⟹TSAcylinder=87.12m2

\therefore\:\underline{\boxed{\green{\sf\:Total\:surface\:area\:of\:cylinder\:=\:87.12\:m^2\:}}}∴Totalsurfaceareaofcylinder=87.12m2

Now, from the given condition,

\displaystyle{\pink{\sf\:Steel\:used\:to\:make\:tank\:-\:Steel\:wasted\:=\:Total\:surface\:area\:of\:cylinder}}Steelusedtomaketank−Steelwasted=Totalsurfaceareaofcylinder

\displaystyle{\implies\sf\:Steel\:used\:-\:\dfrac{1}{12}\:Steel\:used\:=\:TSA_{cylinder}}⟹Steelused−121Steelused=TSAcylinder

\displaystyle{\implies\sf\:Steel\:used\:\left(\:1\:-\:\dfrac{1}{12}\:\right)\:=\:87.12}⟹Steelused(1−121)=87.12

\displaystyle{\implies\sf\:Steel\:used\:\left(\:\dfrac{12\:-\:1}{12}\:\right)\:=\:87.12}⟹Steelused(1212−1)=87.12

\displaystyle{\implies\sf\:Steel\:used\:\times\:\dfrac{11}{12}\:=\:87.12}⟹Steelused×1211=87.12

\displaystyle{\implies\sf\:Steel\:used\:=\:\dfrac{\cancel{87.12}\:\times\:12}{\cancel{11}}}⟹Steelused=

Explanation:

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Answer 2

The focal length of a concave mirror is 25cm

Given,

u = object distance = 50cm

v = image distance = 50cm

To Find,

focal length = f

Solution,

using mirror formula

1/u +1/v= 1/f

= 1/50+1/50 = 2/50

f = 25 cm

to get a diminished image object should be placed at infinity or very far from the mirror.

to obtain the exact size of the image and object, the object must be placed at the center of curvature

So, focal length of the concave mirror is 25cm

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