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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 401. |
A peson looks into a spherical mirror. The size of image of his face is twice the actual size of his face. If the face is at a distance of 20cm, then find the radius of curvature of the mirror. |
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Answer» The person will see his face only when the image is virtual. Virtual image of a real object is erect. Therefore, `m=2` `:.(-v)/(u)=2 (" Here " u = -20 cm)` `rArr v=40cm` Applying `(1)/(4)+(1)/(u)=(1)/(f), f=-40cm` or `R=80 cm`. Aliter : `m=(f)/(f-u)` `rArr 2=(f)/(f-(-20)) rArrf=-40 cm` or `R=80 cm`. |
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| 402. |
The magnifying power of a telescope can be increasedA. by increasing focal lengths of both lengthB. by fiting eyeplace of high powerC. by fiting eye place of low powerD. by increasing the distance of object |
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Answer» Correct Answer - B `MP=-((f_(0))/(f_(theta)))` to increase the magnifying we will decreases `f_(0)` So, we will use eyepiece of high power |
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| 403. |
In a compound microscopeA. the objective has a shorter focal lengthB. the objective has a shorter apertureC. (1) and (2)D. the aperture of objective and eyepiece are same. |
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Answer» Correct Answer - C We use objective of shorter focal length and shorter aperture because the object is placed near the lense. |
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| 404. |
Can a single lens ever form a real and erect image? |
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Answer» For a real object, it is not possible for a single lens to form a real and erect image. But for a virtual object, it may be possible. |
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| 405. |
Under which of the following conditions will a convex mirrorr of focal length `f` produce an image that is erect, diminished and virtual ?A. Only when `2fgtugtf`B. Only when `u=f`C. Only when `u lt f`D. Always |
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Answer» Correct Answer - D Image formed by convex mirrorr is always. Erect diminished and virtual. |
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| 406. |
The table below shows object and image distances for four objects placed in front of mirrors. For which one is the image formed by a convex spherical mirror? [Positive and negative signs are used in accordance with standard sign convention]A. Object distance Image distance `-7.10cm` `-18.0cm`B. Object distance Image distance `-25.0cm` `-16.7cm`C. Object distance Image distance `-5.0cm` `+1.0cm`D. Object distance Image distance `-20.0cm` `+5.71cm` |
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Answer» Correct Answer - d. For (a) and (b), object and image are on the same side, so image is real as the object is real. For (c), image is virtual but `|m|gt1`. So, concave mirror. For (d), `|m|lt1` and image is also virtual, so in (d) image is formed in a convex mirror. |
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| 407. |
Figure. (a)shows an object placed in front of a plane mirror. P,Q and R are the three positions where the image of object may be seen, Observer A is able to see the image at position Q. Where does the observer B see the image of the object? |
| Answer» Observer B also observes the image of the object of the position Q which is explained by the ray diagram in Fig (b). The position of image will be independent of the position of the observer. | |
| 408. |
Two convex lenses placed in contact form the image of a distant object at P. If the lens B is moved to the right, the image will A. move to the leftB. move to the rightC. remain at PD. move either to the left or right, depending upon focal lengths of the lenses. |
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Answer» Correct Answer - b. Power of the system decreases due to separation between the lenses. So, the focal length increases. |
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| 409. |
When the distance between the object and the screen is more than `4 f`. We can obtain image of the object on the screen for the two positions of the lens. If is called displacement method. In one case, the image is magnified and in the other case, it is diminished. If `I_(1)` and `I_(2)` be the sized of the two images, then the size of the object isA. `sqrt(I_(1)I_(2))`B. `sqrt((I_(1))/(I_(2)))`C. `I_(1)-I_(2)`D. `(I_(1)+I_(2))/(2)` |
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Answer» Correct Answer - A `(I_(1))/(O)=(v)/(u),=(u)/(v),(I_(1)I_(2))/(O_(2))=I,O=sqrt(I_(1)I_(2))` |
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| 410. |
A convex lens has a focal length of 10 cm. Find the location and nature of the image if a point object is placed on the principal axis at a distance of (a) 9.8 cm, (b) 10.2 cm from the lens. |
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Answer» Correct Answer - A::B::C::D a. `f=10 cm, u=9.8` 1/v-1/u=1/f` `1/v=1/f+1/u=1/10-1/9.8` `1/v=(9.8-10)/98=(-0.2)/98` `v=-98xx5` ltbgt `=-490cm` (same side of object) `v=490cm` on the side of object (virtual) b. u=10.2 `1/v=1/10-1/10.2` `=(10.2-10)/102=0.2/102` `v=102xx5=510cm(real)` (opposite side of object) |
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| 411. |
A lens forms a virtual, diminished image of an object placed at `2 m` from it. The size of image is half of the object. Which one of the following statements is correct regarding the nature and focal length of the lens ?A. Concave `|f|=1m`B. Convex, `|f|=1`C. Concave,`|f|=2m`D. Convex, `|f|=2m` |
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Answer» Correct Answer - C `m=(f)/(f+u),(1)/(2)=(f)/(f-2)` or `2f=f-2` or `f=-2 metre` `|f|=2 metre` Since the image is virtual as well as diminished therefore the lens is concave. |
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| 412. |
A lens forms a virtual, diminished image of an object placed at 2m from it. The size of image is half of the object. Which one of the following statements is correct regarding the nature and focal length of the lens?A. Concave, `|f|=1m`B. Convex, `|f|=1m`C. Concave, `|f|=2m`D. Convex, `|f|=2m` |
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Answer» Correct Answer - c. `m=(f)/(f+u),(1)/(2)=(f)/(f-2),` or `2f=f-2 ` or `f=-2m` `|f|=2` metre. Since the image is virtual as well as diminshed, therefore the lens is concave. |
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| 413. |
The image of an exend object placed perpendicular to the principal axis of a mirror, will be erect ifA. the object and the image are both realB. the object and the image are both virtualC. the object is real but the image is virtualD. the object is virtual but the image is real |
| Answer» Correct Answer - C::D | |
| 414. |
Shown an object and its image formed by a thin lens. Then the nature and focal length of lens is A. `f=4.8cm` converging lensB. `f=-4.8 cm ` diverging lensC. `f=2.18cm` converging lensD. `f=-2.18cm` diverging lens |
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Answer» Correct Answer - b. `(1)/(v)-(1)/(u)=(1)/(f)` `v=-3cm,u=-8cm` `(1)/(-3)-(1)/(-8)=(-8+3)/(24)=(-5)/(24)` `f=(-24)/(5) cm=-4.8cm` (diverging lens) |
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| 415. |
A 5.0 diopter lens forms a virtual image which is 4 times the object placed perpendicularly on the principal axis of the lens. Find the distance of the object from the lens. |
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Answer» Correct Answer - A::C `h_i=4h_0, h_i/h_0=4` `h_i/h_0=v/(-u)rarr (-v)/u=4, v-4u` `1/v-1/u=1/f` `1//f=P=5, f=1/5m=20cm` `1/(-4u)+1/u=1/20` `3/(4u)=1/20rarr u=60/415cm` |
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| 416. |
The Figure shows the equatorial circle of a glass sphere of radius R having centre at C. The eye of an observer is located in the plane of the circle at a distance R from the surface. A small insect is crawling along the equatorial circle. (a) Calculate the length (L) of the are on the circle where the insect lies where its image is visible to the observer. (b) Calculate the value of L when the eye is brought very close to the sphere. Refractive index of the glass is `mu =(1)/(sqrt(2))` |
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Answer» Correct Answer - (a) `(piR)/(3)` (b) `piR` |
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| 417. |
The diameter of the sun is `1.4 x 10^9m` and its distance from the earth is `1.5 x 10^11 m`. Find the radius of the image of the sun formed by a lens of focal length 20 cm. |
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Answer» Correct Answer - C Here `u=-150x10^1cm` f=20cm Using lens formula `1/v-1/u=1/f` `rarr 1/v=1/20-1/(150x10^11)` `=(750x10^9-1)/(150xx10^11)` `=~=(750xx10^9)/(150x10^11)` `rarr v=(150xx10^11)/(750xx10^9)` `We know magnification `(m)=v/u=h_2/h_1` `rarr h_2-=v/uxxh_1` `=(150xx10^110/(750xx10^9xx150xx10^11)xx0.7xx10^11mm` `=0.93mm` |
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| 418. |
A ball is dropped from a height of 20m above the surface of water in a lake. The refractive index of water is `4//3`. A fish inside the lake , in the line fall of the ball, is looking at the ball. At an instant when the balll is 12.8m above the water surface, the fish sees the speed of ball as .A. `9ms^(-1)`B. `12ms^(-1)`C. `16ms^(-1)`D. `21.33ms^(-1)` |
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Answer» Correct Answer - c. `V_("ball")^(2)=2xx10xx7.2rArrv=12=12ms^(-1)` `X_("image of ball")=(4)/(3)X_("ball")` `V_("image of ball")=(4)/(3)V_("ball")=(4)/(3)xx12=16ms^(-1)` |
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| 419. |
A convex lens produces a double size real image when an object is placed at a distance of 18 cm from it. Where should the object be placed to produce a triple size real image ? |
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Answer» Correct Answer - A::C Given `h_1=2h_0, h_1v/u` `rarr v/u=2` `rarr v=2u=36cm` 1/v-1/u=1/f` 1/36+1/18=1/f` `rarr 1/f=3/36` `rarr f=12 cm` Now,` h_1=3h_0` `h_i/h_0=v/u` `rarrv/u=3 rarr v=3u` `1/(3u)+1/4=1/12 rarr 3u=40` `u=16cm` |
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| 420. |
A convex lens produces a real image `m` times the size of the object. What will be the distance of the object from the lens ?A. `((m+1)/(m))f`B. `(m-1)f`C. `((m-1)/(m))f`D. `(m+1)/(f)` |
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Answer» Correct Answer - A `m=(f)/(f+u)implies-m=(f)/(f+u)impliesu=-((m+1)/(m))f` |
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| 421. |
A convex lens is made up of three different materials as shown in the figure. For a point object placed on its axis, the number of images formed are A. 1B. 5C. 4D. 3 |
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Answer» Correct Answer - D Number of images `=(` Number of materials `)` |
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| 422. |
When an object is placed on the principal axis of a convex lens at two different positins, it produces the images with magnifications `+2` and `-4` respectively. How many times more away from the lens the image will be formed in the second position as compared to the first postion?A. 2B. 4C. 5D. 10 |
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Answer» Correct Answer - c. `m=+2,u=-x` `v=-2x` `-(1)/(2x)+(1)/(x)=(1)/(f)` (i) Similarly, `m=-4, u=-y, and v=+4y` `(1)/(4y)+(1)/(y)=(1)/(f)` (ii) Equating (i) and (ii), `(1)/(2x)=(5)/(4y)` `x=0.4y` or `y =2.5x` Required ratio of images `=4y:2x` `=(4)(2.5):2x=5:1` |
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| 423. |
Find the distance of object placed in the slab of refractive index `mu` from point P of the curved surface of radius R so that image is formed at infinity: A. `((mu-1)R)/(mu)`B. `(muR)/((mu-1))`C. `(R)/((mu-1))`D. `((mu-1)R)/(2mu)` |
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Answer» Correct Answer - b. `(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)` `mu_(2)=1,mu_(1)=mu,u=-v,R=-R` `(1)/(v)-(u)/((-u))=(1-u)/((-R))` `(1)/(v)=(u(mu-1)-muR)/(uR)` `v=(uR)/(u(mu-1))` For `v=oo,u(mu-1)-muR=0` `u=(muR)/((mu-1))` |
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| 424. |
In Figure. , find the position of final image formed. |
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Answer» For convex lens: `v=(uf)/(u+f)=(-15xx10)/(-15+10)=30cm` For concave lense: `u=(30-25)=5cm` `u=(30-25)=5cm` `v=(uf)/(u+f)=(5(-10))/(5-10)=10cm` So, final image is formed 10 cm to the right of the concave lens. |
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| 425. |
What would be the efficiency of the heat engine diagramed as shown below? A. `300%`B. `133%`C. `75%`D. `25%` |
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Answer» Correct Answer - D `eta = (W)/(Q_("supplied")) implies (500)/(2000)` |
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| 426. |
A mark is made on the bottom of a vessel and over this mark, a glass slab to thickness `3.5 cm` and refractive index `(7)/(4)` is placed. Now water (refractive inidex, `(4)/(3))` is poured into the vessel so that the surface of water is `8 cm` above the upper surface of the slab. Looking down of normally through the water, the apparent depth of the mark below the surface of water will be `:`A. `6.33 cm`B. `7.5 cm`C. `8 cm`D. `10 cm` |
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Answer» Correct Answer - C Using the relation apparent depth `=("real depth")/(mu)` , we get apparent depth `=(3.5)/(7//5)+(8)/(4//3)=(2+6)cm=8cm` |
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| 427. |
For a concave mirrorr, if virtual image is formed, the graph between `m` and `u` is of the formA. B. C. D. |
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Answer» Correct Answer - B In concave mirrorr, if virtual images are formed , `u` can have values zero and `f` At `u=0,m=(f)/(f-u)=(f)/(f)=1` At `u=f,m=(f)/(f-u)=-(f)/(-f-(-f))=oo` |
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| 428. |
The graph between `u` and `v` for a convex mirrorr isA. B. C. D. |
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Answer» Correct Answer - A As `u` goes from `0` to `-prop, v` goes from `+0` to `+f` |
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| 429. |
A convex lens of focal length 40 cm is in contact with a concave lens of focal length 25 cm. The power of the combination isA. `-1.5 D`B. `-6.5D`C. `+6.5D`D. `+6.67 D` |
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Answer» Correct Answer - A Focal length of the combination can be calculated as `implies(1)/(F)=(1)/(f_(1))+(1)/(f_(2))implies(1)/(F)=(1)/((+40))+(1)/((-25))` `impliesF=-(200)/(3)cm` `:. P=(100)/(F)=(100)/(-200//3)=-1.5D` |
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| 430. |
A convex lens is placed in contact with a mirrorr as shown. If he space between them is filed with water, its power will A. decreasesB. increasesC. remain unchangedD. increases or decreases depending of the focal length |
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Answer» Correct Answer - A The power of the given system is a combination of the positive powerof the convex lens, negative power of the plane mirrorr. Clearly, the power of the system decreases. |
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| 431. |
If the space between the lenses in the lens combination shown were filled with water, what should happen to the focal length and power of the lens combination ? A. Focal length `=` Decreased, Power`=`increasedB. Focal length `=` Decreased, Power`=` unchangedC. Focal length `=` Increased, Power `=`unchnagedD. Focal length `=` Increased, Power`=` decreased |
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Answer» Correct Answer - D `P=(mu-1)((1)/(R_(1))-(1)/(R_(2)))` `mu` decreaes, `P` decrease. `f` increases. |
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| 432. |
A prism of angle `60^(@)` is made of glass of refractive index 1.50 for red and 1.56 for violet. Find the angular separation of these rays when a narrow pencil of composite light is incident at minimum deviation. |
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Answer» Correct Answer - `5^(@)22` |
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| 433. |
A man washign to get a picture of a Zebra photographed a white donkey after fitting a glass with black streaks onto the objective of his camera.A. the image will look like a white donkey on the photographB. the image will look like a Zebra on the photographC. the image will be more intense compared to the case in which no such glass is usedD. the image will be less intense compared to the case in which no such glass is used |
| Answer» Correct Answer - AD | |
| 434. |
A lens has a power of `+-` diopeters in air. What will be its power if completely immersed in water? `(._amu_(w)=4//3 and _amu_(g)=3//2)` |
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Answer» Let `f_(a)` and `f_(w)` be the focal lengths of the lens in air and water, respectively, then `P_(a)=(1)/(f_(a)) or +-5 =(1)/(f_(a))` `f_(a)=0.2 m =20cm` Now, `(1)/(f_(a))=(._amu_(g)-1)[(1)/(R_(1))-(1)/(R_(2))]` (i) and `(1)/(f_(w))=(._wmu_g-1)[(1)/(R_(1))-(1)/(R_(2))]` (ii) Dividing Eq. (i) by Eq. (ii), we get `(f_(w))/(f_(a))=[(._amu_(g)-1)/(._wmu_(g)-1)]` Again, `._wmu _(g)=(._amu_(g))/(._amu_(w))=(3//2)/(1//8)=(9)/(8)` `rArr(f_(w))/(f_(a))=((3//2)-1)/((9//8)-1)=((1//2))/((1//8))=4` `f_(w)=f_(a)xx5=20xx4=80cm=0.8m` `P_(w)=(1)/(f_(w))=(1)/(0.8)=1.25 diopter. ` |
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| 435. |
A convex lens is held directly above a con lying on a table and forms an image of it. After the lens has been moved vertically a distance equal to its focal length, it forms another image of the coin equal in size with the previous image. If the diameter of the coin is 16 mm, what is the diameter in milimeters of the image? |
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Answer» Magnitude of magnification remain unchanged `m=|(f)/(f+(-x_(0)))|=|(f)/(f+(-x_(0)-f))|=(1)/(f-x_(0))=+-(1)/(x_(0))` `+-x_(0)=f-x_(0)` either `f=0` or `x_(0)=(f)/(2)` `m=(f)/(f-f//2)=2(d_("image"))/(d_("coin"))` `rArrd_("image")=2xxd_("coin")=2xx16=32mm` |
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| 436. |
1 mole of a gas with `gamma=7//5` is mixed with 1 mole of a gas with `gamma=5//3`, then the value of `gamma` for the resulting mixture isA. `7//5`B. `2//5`C. `24//16`D. `12//7` |
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Answer» Correct Answer - C `(n_(1) + n_(2))/(gamma-1) = (n_(1))/(gamma_(1) - 1) + (n_(2))/(gamma_(2) - 1)` `implies (1 + 1)/(gamma-1) = (1)/((5//3) - 1) + (1)/((7//5) - 1)` `implies (2)/(gamma-1) = (3)/(2) + (5)/(2) = 4 implies gamma = (3)/(2) = (24)/(16)` |
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| 437. |
At what temperature is the rms velocity of a hydrogen molecule equal to that of an oxygen molecule at `47^(@)C`?A. `80K`B. `-73K`C. `3K`D. `20K` |
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Answer» Correct Answer - D `V_("rms") = sqrt((3RT)/(M))` Where `T` is the temperature of the gas molecules in kelvin and `M` is the molecules mass of the gas `V_(H_(2))= V_(O_(2)) implies sqrt((T_(H_(2)))/(M_(H_(2)))) = sqrt((T_(O_(2)))/(M_(O_(2)))) implies (T_(H_(2)))/(2) = (320)/(32) implies T_(H_(2)) = 20K` |
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| 438. |
Cooking gas container are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside willA. increaseB. decreaseC. remains sameD. decrease for some , while increase for others |
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Answer» Correct Answer - C The temperature of gas molecules depends on the average kinetic energy associated with the disorderly motion (i.e.,random motion) of the gas molecules . The orderly kinetic energy of the molecules of gas container will increase in the lorry, whereas disorderly kinetic energy will still remain the same , hence the temperature of the gas molecules will remain unchanged. |
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| 439. |
A refrigerator converts `100` g of water at `25^(@)C` into ice at `-10^(@)C` in one hour and `50` minutes. The quantity of heat removed per minute is (specific heat of ice = `0.5 "cal"//g^(@)C`, latent heat of fusion = `80 "cal"//g`)A. `50` calB. `100`calC. `200`calD. `75`cal |
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Answer» Correct Answer - B Heat removed in cooling water from `25^(@)C` to `0^(@)C = 100 xx 1 xx 25 = 2500"cal"` lthrgt Heat removed in converting water into ice at `0^(@)C` = `100 xx80 = 8000 "cal"` Heat removed in cooling ice from `0^(@)C` to `-15^(@)C = 100 xx 0.5 xx 10 = 500 "cal"` Total heat removed in `1` hr `50` min = `2500 + 8000 + 500 = 11000"cal"` Heat removed per minute = `(11000)/(110) = 100 "cal"//"min"` |
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| 440. |
The refractive index of a prism is 2. The prism can have a maximum refracting angle ofA. `90^(@)`B. `60^(@)`C. `45^(@)`D. `30^(@)` |
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Answer» Correct Answer - b. Critical angle `theta_(c)=sin^(-1)((1)/(mu))` `=sin^(-1)((1)/(2))=30^(@)` If `A gt 2theta_(c)` , the ray does not emerge from the prism. So, maximum angle can be `60^(@)` . |
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| 441. |
Angle of minimum deviation is equal to the angle of prism A of an equilateral glass prism. The angle of incidence at which minimum deviation will be obtained isA. `60^(@)`B. `30^(@)`C. `45^(@)`D. `sin^(-1)(2//3)` |
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Answer» Correct Answer - a. `A=delta_(m)=60^(@)` At minimum deviation `i=((A+delta_(m))/(2))=60^(@)` |
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| 442. |
Figure shows a right angled prism ABC having refractive `mu_(g)=(3)/(2)` lowered into water `(mu_(omega)(4)/(3))`. Find angle `alpha` so that the incident ray normal to face AB will be reflected at face BC completely. |
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Answer» Correct Answer - `alpha gt sin^(-1) ((8)/(9))` |
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| 443. |
The plot of deviation `(delta)` vs angle of incidence (i) for a prism is as shown in the figure. Find the angle of the prism (A). |
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Answer» Correct Answer - `32^(@)` |
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| 444. |
For a glass prism the plot of deviation `(delta)` vs angle of incidence (i) is as shown. Find the refractive index of the glass and value of angle `i_(1)`. |
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Answer» Correct Answer - `mu = sqrt(3)` and `i_(1) = 41^(@)` |
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| 445. |
Two identical equilat- eral glass (refractive index `= sqrt(2))` prisms ABC and CDE are kept such that the angle between faces AC and CE is q. A ray of light is incident at an angle i at the face AB and traverses through the two prisms along the path PQRSTU. Find the value of angle i and `theta` such that angle between the incident ray PQ and emergent ray TU is minimum. |
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Answer» Correct Answer - `i=45^(@), theta = 90^(@)` |
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| 446. |
Three glass prism `A,B` and `C` of same refractive index are placed in contact with each other as shown in figure, with no air gap between the prisms. Monochormatic ray of light `OP` passes through the prism assembly and emerges as `QR`. The conditions of minimum deviation is satisfied in the prisms. A. `A` and `C`B. `B` and `C`C. `A` and `B`D. In all prisms `A,B` and `C` |
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Answer» Correct Answer - C In both `A` and `B`, the refracted ray is parallel to the base of prism. |
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| 447. |
One face of a prism with a refrective angle of `30^@` is coated with silver. A ray of light incident on another face at an angle of `45^@` is refracted and reflected from the silver coated face and retraces its path. What is the refractive index of the prism? |
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Answer» Correct Answer - `sqrt(2)` |
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| 448. |
A spherical body of area A and emissivity `e=0.6` is kept inside a perfectly black body. Total heat radiated by the body at temperature TA. `0.60eAT^(4)`B. `0.80eAT^(4)`C. `1.00eAT^(4)`D. `0.40eAT^(4)` |
| Answer» Correct Answer - A | |
| 449. |
An insulated cylinder is divided into three parts A, B and C. Pistons 1 and 2 are connected by a rigid rod and can slide without friction inside the cylinder. Piston 1 is perfectly conducting while piston 2 is perfectly insulating. Equal quantity of an ideal gas is filled in three compartments and the state of gas in every part is same `(P_(0) V_(0) T_(0))`. Adiabatic exponent of the gas is g = 1.5. The compartment B is slowly given heat through a heater H such that the final volume of gas in part C becomes `(4V_(0))/(9)` (a) Calculate the heat supplied by the heater. (b) Calculate the amount of heat flow through piston 1. (c) If heater were in compartment A, instead of B how would your answers to (a) and (b) change? |
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Answer» Correct Answer - (i) `P_(A) = P_(C) = (27)/(8)P_(0) , P_(B) = (21)/(4)P_(0)` (ii) `T_(A) = T_(B) = (21)/(4)T_(0), T_(C) = (3)/(2)T_(0)` (iii) `18P_(0)V_(0) (iv) W_(A) = P_(0)V_(0) , W_(B)= 0` (v)`(17)/(2)P_(0)V_(0)` For compartment `C` `P_(0)V_(0)^(gamma) = P((4V_(0))/(9))^(gamma)` `" "` `implies P = (27)/(8)P_(0)` `implies P_(0)T_(0)^(gamma//1-gamma) = PT^(gamma//1-gamma)` `P_(0)T_(0)^(-3) = ((27)/(8)P_(0)) xx T^(-3) implies T = (3)/(2)T_(0)` For compartment `A` `P_(A) = (27)/(8)P_(0)` `(P_(0)V_(0))/(RT_(0)) = (P_(1)V_(1))/(RT_(1))` `implies T_(1) = ((27)/(8)P_(0)(2V_(0)-(4V_(0))/(9)))/(R) xx (RT_(0))/(P_(0)V_(0)) = T_(1) = (21)/(4)T_(0)` For compartment `B` `(P_(0)V_(0))/(T_(0)) = (P_(1)V_(1))/(T_(1)) implies (P_(0)V_(0))/(T_(0))= (P_(1)V_(0))/(((21)/(4)T_(0)))implies P_(1) = (21)/(4)P_(0)` Final pressure in `A (27)/(8)P_(0)` Final pressure in `B (21)/(4)P_(0)` Final pressure in `C (3)/(2)P_(0)` (ii) Final temperature in `A = (21)/(4) T_(0)` Final temperature in `B = (21)/(4) T_(0)` Final temperature in `C = (3)/(2) T_(0)` (iii) Heat supplied by heater = (`DeltaU + W)` ` Q = (DeltaU_(1) + W_(1)) + (DeltaU_(2) + W_(2)) + 0` =(`DeltaU_(1) - W_(3)) + (DeltaU_(2) + 0)` =`DeltaU_(1) + DeltaU_(2) - W_(3)` = ` (n_(1)RDeltaT)/(gamma_(1) - 1) + (n_(2)RDeltaT)/(gamma_(2)-1) - ((n_(3) RDeltaT)/(gamma_(3)-1))` =`(17)/(2)P_(0)V_(0) + (17)/(2)P_(0)V_(0) + P_(0)V_(0) = 18 P_(0)V_(0)` (iv) Work done by gas in chamber `B =0` Work done by gas in chamber `C = -DeltaU = (-nRDeltaT)/(gamma-1) = -P_(0)V_(0)` Work done by gas in chamber `A = (-) W_(chamber) = -(P_(0)V_(0)) = P_(0)V_(0)` (v) Heat flowing across piston-I = `DeltaU_(2)= (17)/(2)P_(0)V_(0)` |
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A refrigerator takes heat from water at `0^(@)C` and transfer it to room at `27^(@)C` . If `100 kg` of water is converted in ice at `0^(@)C` then calculate the work done. (Latent heat of ice `3.4xx10^(5) J//kg`) |
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Answer» Coefficient of performance `(COP) = (T_(2))/(T_(1)-T_(2)) = (273)/(300-273) = (273)/(27)` ` W= (Q_(2))/(COP) = (mL)/(COP) = (100xx 3.4xx10^(5))/(273//27) = (100 xx 3.4 xx 10^(5) xx27)/(273) = 3.36 xx 10^(7)J` |
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