Class 10th Science - Light Reflection and Refraction Case Study Questions and Answers 2022 - 2023
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Light Reflection and Refraction Case Study Questions With Answer Key
10th Standard CBSE
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Reg.No. :
Science
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The curved surface of a spoon can be considered as a spherical mirror. A highly smooth polished surface is called mirror. The mirror whose reflecting surface is curved inwards or outwards is called a spherical mirror. Inner part works as a concave mirror and the outer bulging part acts as a convex mirror. The center of the reflecting surface of a mirror is called pole and the radius of the sphere of which the mirror is formed is called radius of curvature.
(i) When a concave mirror is held towards the sun and its sharp image is formed on a piece of carbon paper for some time, a hole is burnt in the carbon paper. What is the name given to the distance between the mirror and carbon paper?(a) Radius of curvature (b) Focal length (c) Principal focus (d) Principal axis (ii) The distance between pole and focal point of a spherical mirror is equal to the distance between
(a) pole and center of curvature (b) focus point and center of curvature (c) pole and object (d) object and image (iii) The focal length of a mirror is 15 cm. The radius of curvature is
(a) 15 cm (b) 30 cm (c) 45 cm (d) 60 cm (iv) The normal at any point on the mirror passes through
(a) focus (b) pole (c) center of curvature (d) any point (v) In a convex spherical mirror, reflection of light takes place at
(a) a flat surface (b) a bent-in surface (c) a bulging-out surface (d) an uneven surface -
The spherical mirror forms different types of images when the object is placed at different locations.
When the image is formed on screen, the image is real and when the image does not form on screen, the image is virtual. When the two reflected rays meet actually, the image is real and when they appear to meet, the image is virtual.
A concave mirror always forms a real and inverted image for different positions of the object. But if the object is placed between the focus and pole, the image formed is virtual and erect.
A convex mirror always forms a virtual, erect and diminished image. A concave mirror is used as doctor's head mirror to focus light on body parts like eyes, ears, nose etc., to be examined because it can form erect and magnified image of the object. The convex mirror is used as a rear view mirrors in automobiles because it can form an small and erect image of an object.
(i) When an object is placed at the centre of curvature of a concave mirror, the image formed is(a) larger than the object (b) smaller than the object (c) same size as that of the object (d) highly enlarged. (ii) No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane (b) concave (c) convex (d) either plane or convex. (iii) A child is standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body of the same size and that of the legs smaller. The following is the order of combinations for the magic mirror from the top.
(a) Plane, convex and concave (b) Convex, concave and plane (c) Concave, plane and convex (d) Convex, plane and concave (iv) To get an image larger than the object, one can use
(a) convex mirror but not a concave mirror
(b) a concave mirror but not a convex mirror
(c) either a convex mirror or a concave mirror
(d) a plane mirror.
(v) A convex mirror has wider field of view because
(a) the image formed is much smaller than the object and large number of images can be seen
(b) the image formed is much closer to the mirror
(c) both (a) and (b)
(d) none of these. -
The relation between distance of an object from the mirror (u), distance of image from the mirror (v) and the focal length (F) is called mirror formula. This formula is valid in all situations for all spherical mirrors for all positions of the object. The size of image formed by a spherical mirror depends on the position of the object from the mirror. The image formed by a spherical mirror can be bigger than the object, equal to the object or smaller than the object. The size of the image relative to the object is given by the linear magnification (m). Thus, the magnification is given by the ratio of height of image to the height of object. If magnification is negative, image is real and if it is positive, image is virtual.
(i) What is the position of an image when an object is placed at a distance of 20 em from a concave mirror of
focal length 20 cm?(a) 5 cm (b) 20 cm (c) 10 cm (d) infinity (ii) Which of the following ray diagrams is correct for the ray of light incident on a concave mirror as shown in figure?
(a) Figure A
(b) Figure B
(C) Figure C
(d) Figure D(iii) If the magnification of an image is -2, the characteristic of image will be
(a) real and inverted (b) virtual and enlarged (c) virtual and inverted (d) real and small (iv) The mirror formula holds for
(a) concave mirror (b) convex mirror (c) plane mirror (d) all of these (v) A parallel beam of light is made to fall on a concave mirror. An image is formed at a distance of7.5 from the mirror. The focal length of the mirror is
(a) 15 cm (b) 7.5 cm (c) 3.75 cm (d) 10 cm -
When the rays of light travels from one transparent medium to another, the path of light is deviated. This phenomena is called refraction of light. The bending of light depends on the optical density of medium through which the light pass.
The speed of light varies from medium to medium. A medium in which the speed of light is more is optically rarer medium whereas in which the speed of light is less is optically denser medium. Whenever light goes from one medium to another, the frequency of light does not change however, speed and wavelength change. It concluded that change in speed of light is the basic cause of refraction.
(i) When light travels from air to glass, the ray of light bends(a) towards the normal (b) away from normal (c) anywhere (d) none of these (ii) A ray of light passes from a medium A to another medium B. No bending of light occurs if the ray of light hits the boundary of medium B at an angle of
(a) 0° (b) 45° (c) 90° (d) 120° (iii) When light passes from one medium to another, the frequency of light
(a) increases (b) decreases (c) remains same (d) none of these (iv) When light passes from glass to water, the speed of light
(a) increases (b) decreases (c) remains same (d) first increases then decrease (v) The bottom of pool filled with water appears to be ______ due to refraction of light
(a) shallower (b) deeper (c) at same depth (d) empty -
The refraction of light on going from one medium to another takes place according to two laws which are known as the laws of refraction of light. These laws are
1. The ratio of sine of angle of incidence to the sine of angle of refraction is always constant for the pair of media in contact.
\(\frac{\sin i}{\sin r}=\mu=\text { constant }\)
This constant is called refractive index of the second medium with respect to the first medium.
Refractive index is also defined as the ratio of speed of light in vacuum to the speed of light in medium.
2. The incident ray, refracted ray and normal all lie in the same plane.
This law is called Snell's law of refraction.
(i) When light travels from air to glass,
(a) angle of incidence > angle of refraction
(b) angle of incidence < angle of refraction
(c) angle of incidence = angle of refraction
(d) can't say
(ii) When light travels from air to medium, the angle of incidence is 45° and angle of refraction is 30°. The refractive index of second medium with respect to the first medium is(a) 1.41 (b) 1.50 (c) 1.23 (d) 1 (iii) In which medium, the speed of light is minimum?
(a) Air (b) Glass (c) Water (d) Diamond (iv) If the refractive index of glass is 1.5 and speed of light in air is 3 x 108 m/s. The speed of light in glass is
(a) 2 x 108 mls (b) 2.9 x 108 mls (c) 4.5 x 108 mls (d) 3 x 108 mls (v) Refractive index of a with respect to b is 2. Find the refractive index of b with respect to a.
(a) 0.4 (b) 0.5 (c) 0.25 (d) 2. -
A lens is a piece of any transparent material bounded by two curved surfaces. There are two types of lenses convex lens and concave lens.
Convex lens is made up of a transparent medium bounded by two spherical surfaces such that thicker at the middle and thinner at the edges. Concave lens is also made up of a transparent medium such that thicker at the edge and thinner at the middle. The mid point of the lens is called optical centre.
A point on the principal axis, where the incident parallel rays meet or appears to come out after refraction is called focus.
A convex lens converges a parallel beam of light to other side whereas concave lens spreads out.
(i) Which of the following lenses would you prefer to use while reading small letters found in dictionary?
(a) A convex lens of focal length 50 cm
(b) A concave lens of focal length 50 cm
(c) A convex lens of focal length 5 cm
(d) A concave lens of focal length 5 cm
(ii) Which type of lenes are shown in given figure (i) and (ii).
(a) Plano concave, concavo convex
(b) Plano convex, convexo concave
(c) Double concave, concave convex
(d) Convexo concave, double convex
(iii) A small bulb is placed at the focal point of a converging lens. When the bulb is switched on, the lens produces
(a) a convergent beam of light
(b) a divergent beam of light
(c) a parallel beam of light
(d) a patch of coloured light
(iv) The part oflens through which the refraction takes place is called(a) aperture (b) centre of curvature (c) principal axis (d) focus (v) A water drop acts as a
(a) convex lens (b) concave lens (c) double concave lens (d) none of these -
The lenses forms different types of images when object placed at different locations. When a ray is incident parallel to the principal axis, then after refraction, it passes through the focus or appears to come from the focus. When a ray goes through the optical centre of the lens, it passes without any deviation.
If the object is placed between focus and optical center of the convex lens, erect and magnified image is formed. As the object is brought closer to the convex lens from infinity to focus, the image moves away from the convex lens from focus to infinity. Also the size of image goes on increasing and the image is always real and inverted. A concave lens always gives a virtual, erect and diminished image irrespective to the position of the object.
(i) The location of image formed by a convex lens when the object is placed at infinity is(a) at focus (b) at 2F (c) at optical center (d) between F and 2F (ii) When the object is placed at the focus of concave lens, the image formed is
(a) real and smaller (b) virtual and inverted (c) virtual and smaller (d) real and erect (iii) The size of image formed by a convex lens when the object is placed at the focus of convex lens is
(a) small (b) point in size (c) highly magnified (d) same as that of object (iv) When the object is placed at 2F in front of convex lens, the location of image is
(a) at F (b) at 2 F on the other side (c) at infinity (d) between F and optical center -
The relationship between the distance of object from the lens (u), distance of image from the lens (v) and the focal length (j) of the lens is called lens formula. It can be written as\(\begin{equation} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \end{equation}\)
The size of image formed by a lens depends on the position of the object from the lens. A lens of short focal length has more power whereas a lens of long focal length has less power. When the lens is convex, the power is positive and for concave lens, the power is negative.
The magnification produced by a lens is the ratio of height of image to the height of object as the size of the image relative to the object is given by linear magnification (m).
When, m is negative, image formed is real and when m is positive, image formed is virtual. If m < 1, size of image is smaller than the object. If m > 1, size of image is larger than the object.
(i) An object 4 cm in height is placed at a distance of 10 cm from a convex lens of focal length 20 cm. The position of image is(a) - 20 cm (b) 20 cm (c) -10 cm (d) 10 cm (ii) In the above question, the size of image is
(a) 16 cm (b) 8 cm (c) 4 cm (d) 2 cm (iii) An object is ,placed 50 cm from a concave lens and produces a virtual image at a distance of 10 cm in front of lens. The focal length of lens is
(a) - 25 cm (b) -12.5 cm (c) 12.5 cm (d) 10 cm (iv) A convex lens forms an image of magnification -2 of the height of image is 6 cm, the height of object is
(a) 6 cm (b) 4 cm (c) 3 cm (d) 2 cm (v) A concave lens of focal length 5 cm, the power of lens is
(a) 20D (b) -20D (c) 90D (d) -5 D -
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. She is pulling the string at the rate of 5 cm per second. Nazima’s friend observe her position and draw a rough sketch by using A, B, C and D positions of tip, point directly under the tip of the rod, fish and Nazima’s position (see the below figure). Assuming that her string (from the tip of her rod to the fly) is taut, answer the following questions:
(i) What is the length AC?(a) 2 m (b) 3 m (c) 4 m (d) 5 m (ii) What is the length of string pulled in 12 seconds?
(a) 6 m (b) 0.3 m (c) 0.6 m (d) 3 m (iii) What is the length of string after 12 seconds?
(a) 2.4 m (b) 2.7 m (c) 2 m (d) 2.2 m (iv) What will be the horizontal distance of the fly from her after 12 seconds?
(a) 2.7 m (b) 2.78 m (c) 2.58 m (d) 2.2 m (v) The given problem is based on which concept?
(a)Triangles (b) Co-ordinate geometry (c) Height and Distance (d) None of these -
Junk food is unhealthful food that is high in calories from sugar or fat, with little dietary fiber, protein, vitamins, minerals, or other important forms of nutritional value. A sample of few students have taken. If α be the number of students who take junk food, β be the number of students who take healthy food such that \(\alpha>\beta \text { and } \alpha \text { and } \beta\) and are the zeroes of the quadratic polynomial f(x) = x2 – 7x + 10, then answer the following questions:
(i) Name the type of expression of the polynomial in the above statement?(a) quadratic (b) cubic (c) linear (d) bi-quadratic (ii) Find the number of students who take junk food.
(a) 5 (b) 2 (c) 7 (d) None of these (iii) Find the number of students who take healthy food.
(a) 5 (b) 2 (c) 7 (d) None of these (iv) Find the quadratic polynomial whose zeros are -3 and -4.
(a) x2 + 4x + 2 (b) x2 – x – 12 (c) x2 – 7x + 12 (d) None of these (v) If one zero of the polynomial x2 – 5x \(+\) 6 is 2 then find the other zero.
(a) 6 (b) -6 (c) 2 (d) None of these -
Mathematics teacher of a school took her 10th standard students to show Red fort. It was a part of their Educational trip. The teacher had interest in history as well. She narrated the facts of Red fort to students. Then the teacher said in this monument one can find combination of solid figures. There are 2 pillars which are cylindrical in shape. Also 2 domes at the corners which are hemispherical.7 smaller domes at the centre. Flag hoisting ceremony on Independence Day takes place near these domes.
(i) How much cloth material will be required to cover 2 big domes each of radius 2.5 metres? (Take π = 22/7)(a) 75 m2 (b) 78.57 m2 (c) 87.47 m2 (d) 25.8 m2 (ii) Write the formula to find the volume of a cylindrical pillar :
(a) \(\pi\)r2h (b) \(\pi\)rl (c) \(\pi\)r (l+r) (d) 2\(\pi\)r (iii) Find the lateral surface area of two pillars if height of the pillar is 7 m and radius of the base is 1.4m.
(a) 112.3 cm2 (b) 123.2 m2 (c) 90 m2 (d) 345.2 cm2 (iv) How much is the volume of a hemisphere if the radius of the base is 3.5 m?
(a) 85.9 m3 (b) 80 m3 (c) 98 m3 (d) 89.83 m3 (v) What is the ratio of sum of volumes of two hemispheres of radius 1 cm each to the volume of a sphere of radius 2 cm?
(a) 1:1 (b) 1:8 (c) 8:1 (d) 1:16 -
On one day, a poor girl of height 90 cm is looking for a lamp-post for completing her homework as in her area power is not there and she finds the same at some distance away from her home. After completing the homework, she is walking away from the base of a lamp-post at a speed of 1.2 m/s. The lamp is 3.6 m above the ground (see below figure).
(i) Find her distance from the base of the lamp post.(a) 1.2 m (b) 3.6 m (c) 4.8 m (d) none of these (ii) Find the correct similarity criteria applicable for triangles ABE and CDE.
(a) AA (b) SAS (c) SSS (d) AAS (iii) Find the length of her shadow after 4 seconds.
(a) 1.2 m (b) 3.6 m (c) 4.8 m (d) none of these (iv) Sides of two similar triangles are in the ratio 9:16. Find the ratio of Corresponding areas of these triangles.
(a) 9:16 (b) 3:4 c) 81:256 (d) 18:32 (v) Find the ratio AC:CE.
(a) 1: 3 (b) 3 : 1 (c) 1 : 4 (d) 4 : 1 -
In the below given layout, the design and measurements has been made such that area of two bedrooms and Kitchen together is 95 sq. m.
(i) The area of two bedrooms and kitchen are respectively equal to(a) 5x, 5y (b) 10x, 5y (c) 5x, 10y (d) x, y (ii) Find the length of the outer boundary of the layout.
(a) 27 m (b) 15 m (c) 50 m (d) 54 m (iii) Find the area of each bedroom.
(a) 30 sq. m (b) 35 sq. m (c) 65 sq. m (d) 42 sq. m (iv) Find the area of living room in the layout.
(a) 30 sq. m (b) 35 sq. m (c) 75 sq. m (d) 65 sq. m (v) Find the cost of laying tiles in Kitchen at the rate of Rs. 50 per sq. m
(a) Rs. 1500 (b) Rs. 2000 (c) Rs. 1750 (d) Rs. 3000 -
A student made a wooden pen stand which is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see the below figure)
(i) What is the volume of cuboid?(a) 525 cm3 (b) 225 cm3 (iii) 552 cm3 (d) 255 cm3 (ii) What is the volume of cone?
(a) \( \frac{11}{3} \mathrm{~cm}^{3}\) (b)\(\frac{11}{30} \mathrm{~cm}^{3}\) (c)\(\frac{3}{11} \mathrm{~cm}^{3} \) (d)\(\frac{30}{11} \mathrm{~cm}^{3}\) (iii) What is the total volume of conical depressions?
(a) 1.74 cm3 (b) 1.44 cm3 (c) 1.47 cm3 (d) 1.77 cm3 (iv) What is the volume of wood in the entire stand?
(a) 522.35 cm3 (b) 532.53 cm3 (c) 523.35 cm3 (d) 523.53 cm3 (v) The given problem is based on which mathematical concept?
(a) Triangle (b) Surface Areas & volumes (c) Height & Distances (d) None of these -
In a potato race, a bucket is placed at the starting point, which is 4 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see below figure).
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket.
(i) What is the distance covered by the competitor in first potato?(a) 10 m (b) 8 m (c) 12 m (d) 14m (ii) What is the distance covered by the competitor in second potato?
(a) 14 m (b) 12 m (c) 10 m (d) 8 m (iii) What is the distance covered by the competitor in fourth potato?
(a) 22 m (b) 24 m (c) 26 m (d) 28 m (iv) What is the total distance covered by the competitor in first and second potato?
(a) 22 m (b) 24 m (c) 26 m (d) 30 m (v) If the A.P. 8, 14, 20, ..., then find the common difference.
(a) 4 (b) 8 (c) 12 (d) 6
Case Study
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Answers
Light Reflection and Refraction Case Study Questions With Answer Key Answer Keys
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(i) (b): The focal length of a concave mirror is the distance between its pole and principal focus.
(ii) (b)
(iii) (b): Given that, f = 15 cm
Radius of curvature of a spherical mirror is given as
R= 2F
R = 2 x 15 = 30 cm.
(iv) (c): In a spherical mirror, normal drawn at any point passes through the centre of curvature
(v) (c) -
(i) (c):
When the object is placed at the centre of curvature of concave mirror, the image formed is real, inverted and of the same size as that of the object.
(ii) (d): The image is erect in a plane mirror and also in a convex mirror, for all positions of the object.
(iii) (c) : As the image of head is bigger, the upper portion of magic mirror is concave. The middle portion of the image is of same size, so, middle portion of magic mirror is plane. Now, the image of legs looks smaller, therefore, the lower portion of magic mirror is convex.
(iv) (b)
(v) (c) -
(i) (d): When an object is placed at the focus of a concave mirror, the image is formed at infinity.
(ii) (d): When a light ray parallel to the principal axis is incident on a concave mirror, it passes through the principal focus after reflection. Therefore, figure D is correct.
(iii) (a) : If m is negative, the image will be real and inverted.
(iv) (d)
(v) (b): The distance of object from mirror = \(\infty\)
Using, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
\(\frac{1}{\infty}-\left(-\frac{1}{7.5}\right)=\frac{1}{f}\)
f = 7.5 cm -
(i) (a): When, a ray of light travels from air to glass, it bends towards the normal.
(ii) (c): No bending of light occurs when light is incident normally or perpendicularly on a boundary of two media since angle of incidence and angle of refraction both are zero.
(iii) (c): When light goes from one medium to other medium, its frequency does not change
(iv) (a): The speed to light increases when light passes from glass to water as water is optically rarer medium.
(v) (a): The bottom of a pool of water appears to be less deep than it actually is due to refraction. -
(i) (a): According to Snell's law of refraction,
\(\frac{\sin i}{\sin r}>1 \text { or } \sin i>\sin r\)
or i > r.
(ii) (a:) As, 1 \(\mu^{2}=\frac{\sin i}{\sin r}\)
\(\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\frac{1 / \sqrt{2}}{1 / 2}=1.41\)
(iii) (d): As diamond has maximum value of refractive index, therefore it has minimum speed of light in medium.
(iv) (a): As, \(\mu_{\text {glass }}=1.5, c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\)
\(\begin{equation} \begin{array}{l} \mu=\frac{c}{v} \text { or } 1.5=\frac{3 \times 10^{8}}{v} \\ v=2 \times 10^{8} \mathrm{~m} / \mathrm{s} \end{array} \end{equation}\)
(v) (b): Given, refractive index of a with respect to b is \(\begin{equation} { }^{b} \mu_{a}=2 \end{equation}\)
Refractive index of b with respect to a is
\(\begin{equation} \frac{1}{b_{\mu_{a}}}={ }^{a} \mu_{b}=\frac{1}{2}=0.5 \end{equation}\) -
(i) (c): Convex lens is used as magnifying glass.
For better performance its focal length should be small.
(ii) (a)
(iii) (c)
(iv) (a): A aperture is the area of the lens available for refraction.
(v) (a): Water droplets behave like a convex lens only as refraction takes place on outer surface. -
(i) (a): When an object is placed at infinity of convex lens, image will be formed at focus F.
(ii) (b): Virtual and inverted image is formed, when object is placed at focus of the concave lens.
(iii) (c) : When object is placed at focus of a convex lens, highly enlarged or magnified image is formed.
(iv) (b): When an object is placed at distance 2F in front of a convex lens, then the image formed is at a distance 2F on the other of the lens.
(v) (a): Image if formed between focus and optical centre when the object is placed anywhere between optical centre and infinity. -
(i) (a):Given, f= 20 cm, U = -10 cm
Using, \(\begin{equation} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \end{equation}\)
\(\begin{equation} \frac{1}{20}=\frac{1}{v}-\left(-\frac{1}{10}\right) \Rightarrow v=-20 \mathrm{~cm} \end{equation}\)
(ii) (b): As,\(\begin{equation} m=\frac{v}{u}=\left(\frac{-20}{-10}\right)=2 \end{equation}\)
\(\begin{equation} \begin{array}{l} m=\frac{h_{2}}{h_{1}} \\ 2=\frac{h_{2}}{4} \Rightarrow h_{2}=8 \mathrm{~cm} \end{array} \end{equation}\)
(iii) (b): Here u = -50 cm, v = 10 cm,f=?
Using, \(\begin{equation} \frac{1}{f}=\frac{1}{10}-\frac{1}{50} \Rightarrow f=-12.5 \mathrm{~cm} \end{equation}\)
(iv) (c): Here, m = - 2
h2 = - 6 cm
h1 =?
As,\(\begin{equation} m=\frac{h_{2}}{h_{1}} \Rightarrow-2=\frac{-6}{h_{2}} \Rightarrow h_{1}=3 \mathrm{~cm} \end{equation}\)
(v) (b):As \(\begin{equation} P=\frac{1}{f}(\because f=5 \mathrm{~cm}) \end{equation}\)
\(\begin{equation} P=\frac{-1}{0.05 \mathrm{~m}}=-20 \mathrm{D} \end{equation}\) -
(i) (b) 3 cm
AC2 = AB2 + BC2 [ By Pythagoras theorem ]
⇒ AC2 = (1.8)2+ (2.4)2
⇒ AC2 = 3.24 + 5.76
⇒ AC2 = 9
⇒ AC = 3m
(ii) (c) 0.6m
She pulls the string at the rate of 5cm/s
∴ String pulled in 12 second = 12 × 5 = 60cm
= 0.6m
(iii) (a) 2.4m
Length of string out after 12 second is AP.
⇒ AP = AC – String pulled by Nazima in 12 seconds.
⇒ AP = (3 − 0.6)m
=2.4m
(iv) (b) 2.78m
In â–³ADB, AB2 + BP2=AP2 ⇒ (1.8)2 + BP2= (2.4)2
⇒ BP2 = 5.76 − 3.24 ⇒ BP2= 5.76 − 3.24
⇒ BP2 = 2.52 ⇒ BP=1.58 m
Horizontal distance of fly = BP + 1.2m
Horizontal distance of fly =1.58m + 1.2m
∴ Horizontal distance of fly = 2.78m
(v) (a) Triangles -
(i) (a) quadratic
(ii) (d) None of these
Now, f(x) = x2 – 7x + 10 = 0
⇒ x2 – 5x – 2x + 10 = 0
⇒ x(x – 5) – 2(x – 5) = 0
⇒(x – 5)(x – 2) = 0
⇒ x = 5, 2
(iii) (d) None of these
(iv) (d) None of these
Sum of Zeroes = -3- 4= -7, product of Zeroes= -34 x -12
x2 - (Sum of Zeroes)x + (Product of Zeroes)= x2 - (-7) x +12= x2 + 7x +12
(v) (d) None of these
\(\text { Sum of Zeroes }=a+2=\frac{-(-5)}{1}=5 \quad \Rightarrow a=5-2=3\) -
(i) (b) 78.57 m2
Radius of a dome, r = 2.5 cm
The dome is hemispherical in shape.
Then, cloth material required = 2 x Surface area of hemisphere
=2 x 2πr2
\(=4 \times \frac{22}{7} \times 2.5 \times 2.5\) = 78.57 m2
(ii) (a) \(\pi\)r2h
(iii) (b) 123.2 m2
Height of each pillar, h= 7m
Radius of base, r= 1.4 m
Lateral surface area or curved surface area of 2 pillars = 2 x 2\(\pi\)rh
\(=4 \times \frac{22}{7} \times 1.4 \times 7\)
= 123.2 m2
(iv) (d) 89.83 m3
Radius of hemisphere, r= 3.5m
Then, volume of a hemisphere,\(\mathrm{V}=\frac{2}{3} \pi r^{3}\)
\(=\frac{2}{3} \times \frac{22}{7} \times(3.5)^{3}\)
= 89.83 m2
(v) (b) 1:8
Volume of 2 hemispheres of radius \(1 \mathrm{~cm}=2 \times \frac{2}{3} \pi r^{3}=\frac{4}{3} \pi(1)^{3}=\frac{4}{3} \pi \mathrm{cm}^{3}\)
Volume of 1 sphere of radius \(2 \mathrm{~cm}=\frac{4}{3} \pi\left(r^{3}\right)=\frac{4}{3} \pi(2)^{3}=\frac{32}{3} \pi \mathrm{cm}^{3}\)
Then, required ratio \(=\frac{\frac{4}{3} \pi}{\frac{32}{3} \pi}=\frac{1}{8}\) -
(i) (c) 4.8 m
Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post.
From the figure, DE is the shadow of the girl. Let DE be x metres.
Now, her distance from the base of the lamp = BD = 1.2 m × 4 = 4.8 m.
(ii) (a) AA
In Δ ABE and Δ CDE, \(\angle \mathrm{B}=\angle \mathrm{D}\) (Each is of 90%)
and\(\angle \mathrm{E}=\angle \mathrm{E}\) (Same angle)
So,\(\triangle \mathrm{ABE} \sim \Delta \mathrm{CDE}\) (AA similarity criterion)
(iii) (d) none of these
\(\Delta \mathrm{ABF} \sim \Delta \mathrm{CDF} \)
\(\Rightarrow \frac{\mathrm{BE}}{\mathrm{DF}}=\frac{\mathrm{AB}}{\mathrm{CD}} \)
\(\Rightarrow \frac{4.8+x}{x}=\frac{3.6}{0.9} \Rightarrow 4.8+x=4 x \)
\(\Rightarrow \ 3 x=4.8 \ \Rightarrow \ x=1.6\)
So, the shadow of the girl after walking for 4 seconds is 1.6 m long.
(iv) (b) 3:4
Since ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides,
Ratio of areas of similar triangles =\(\sqrt{9}: \sqrt{1} 6\)
= 3 : 4
(v) (b) 3 : 1
\(\frac{A E}{C E}=\frac{B E}{D E}=\frac{4.8+1.6}{1.6}=\frac{6.4}{1.6}=4 \)
\(\Rightarrow A E=4 C E \)
\( \Rightarrow A C+C E=4 C E\)
\(\Rightarrow \quad 3 x=4.8 \ \Rightarrow \ x=1.6\) -
(i) (a) 5x, 5y
Area of one bedroom = 5x sq.m
Area of two bedrooms = 10x sq.m
Area of kitchen = 5y sq. m
(ii) (d) 54 m
Length of outer boundary = 12 + 15 + 12 + 15 = 54 m
(iii) (a) 30 sq. m
Area of two bedrooms = 10x sq.m
Area of kitchen = 5y sq. m
So, 10x + 5y = 95 ⇒ 2x + y = 19
Also, x + 2 + y = 15⇒ x + y = 13
Solving 2x + y = 19 and x + y = 13,
we get x = 6 m and y = 7 m
Area of bedroom = 5 x 6 = 30 sq. m
(iv) (a) 75 sq. m
Area of living room = (15 x 7) – 30
= 105 – 30 =75 sq. m
(v) (c) Rs.1750
Cost of 1m2 laying tiles in kitchen = Rs. 50
Total cost of laying tiles in kitchen = Rs. 50 x 35 = Rs. 1750 -
(i) (a) 525 cm3
Volume of cubiod = lbh
= 15 x 10 x 3.5
=15 x 35 = 525 cm3
(ii) (b) \(\frac{11}{30} \mathrm{~cm}^{3}\)
volume of cone =\(\frac{1}{3} \pi \mathrm{R}^{2} \mathrm{H}=\frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4\)
\(=\frac{1}{3} \times \frac{22}{7} \times \frac{1}{2} \times \frac{1}{2} \times \frac{14}{10}=\frac{11}{30} \mathrm{~cm}^{3}\)
(iii) (c) 1.47 cm3
Volume of 4 cones = 4 x \(\frac{11}{30} \mathrm{~cm}^{3}=\frac{22}{15} \mathrm{~cm}^{3}\) = 1.47 cm3
(iv) (d) 523.53 cm3
Volume of wood = 525 cm3 - 1.47 cm3
=523.53 cm3
(v) (b) Surface Areas & volumes -
(i) (b) 8 m
Distance covered by the competitor in first potato = 2 x 4
= 8 m.
(ii) (a) 14 m
Distance covered by the competitor in second potato = 2 x (4 + 3 x 1)
= 2 x 7 = 14 m.
(iii) (c) 26 m
Distance covered by the competitor in fourth pick = 2 x (4 + 3 x 3)
= 2 x 13 = 26 m.
(iv) (a) 22 m
Total distance covered by the competitor in first and second pick = (8 + 14) m
= 22 m.
(v) (d) 6 m
Common difference = d2 — d1
= 14 – 8 = 6.
Case Study
