Q: Question 1(vi)Which of the following are isobars ?612C^{12}_6\text{C}612C and 614C^{14}_6\text{C}614C816O^{16}_8\text{O}816O and 1123Na^{23}_{11}\text{Na}1123Na1123Na^{23}_{11}\text{Na}1123Na and 1122Mg^{22}_{11}\text{Mg}1122Mg1123Na^{23}_{11}\text{Na}1123Na and 1223Mg^{23}_{12}\text{Mg}1223Mg

1123Na^{23}_{11}\text{Na}1123Na and 1223Mg^{23}_{12}\text{Mg}1223MgReason — Isobars are atoms having the same mass number but different atomic number. Hence, 1123Na^{23}_{11}\text{Na}1123Na and 1223Mg^{23}_{12}\text{Mg}1223Mg are isobars.

Q: Question 1(xv)If refractive index of glass w.r.t. air is 32\dfrac{3}{2}23. What is the refractive index of air w.r.t. glass?34\dfrac{3}{4}4343\dfrac{4}{3}3432\dfrac{3}{2}2323\dfrac{2}{3}32

23\dfrac{2}{3}32Reason — The refractive index of glass w.r.t. air is aμg = 32\dfrac{3}{2}23We know that:gμa=1aμg_{\text{g}}\text{μ}_{\text{a}} = \dfrac{1}{_a\mu_g} \\[1em]gμa=aμg1So, substituting we get,gμa=132=23_{\text{g}}\text{μ}_{\text{a}} = \dfrac{1}{\dfrac{3}{2}} = \dfrac{2}{3} \\[1em]gμa=231=32Hence, gμa = 23\dfrac{2}{3}32

Question 1

Question 1(vi)Which of the following are isobars ?612C^{12}_6	ext{C}612​C and 614C^{14}_6	ext{C}614​C816O^{16}_8	ext{O}816​O and 1123Na^{23}_{11}	ext{Na}1123​Na1123Na^{23}_{11}	ext{Na}1123​Na and 1122Mg^{22}_{11}	ext{Mg}1122​Mg1123Na^{23}_{11}	ext{Na}1123​Na and 1223Mg^{23}_{12}	ext{Mg}1223​Mg

Accepted Answer

1123Na^{23}_{11}	ext{Na}1123​Na and 1223Mg^{23}_{12}	ext{Mg}1223​MgReason — Isobars are atoms having the same mass number but different atomic number. Hence, 1123Na^{23}_{11}	ext{Na}1123​Na and 1223Mg^{23}_{12}	ext{Mg}1223​Mg are isobars.

Question 2

Question 1(xii)A ray is made to fall on a prism. Maximum deviation it can suffer isπ3\dfrac{π}{3}3π​π4\dfrac{π}{4}4π​π2\dfrac{π}{2}2π​πππ

Accepted Answer

πππReason — The maximum deviation produced by a prism is πππ or 180°, when the ray of light enters the prism at an angle of 90° as shown below.

Question 3

Question 1(xv)If refractive index of glass w.r.t. air is 32\dfrac{3}{2}23​. What is the refractive index of air w.r.t. glass?34\dfrac{3}{4}43​43\dfrac{4}{3}34​32\dfrac{3}{2}23​23\dfrac{2}{3}32​

Accepted Answer

23\dfrac{2}{3}32​Reason — The refractive index of glass w.r.t. air is aμg = 32\dfrac{3}{2}23​We know that:gμa=1aμg_{\text{g}}\text{μ}_{\text{a}} = \dfrac{1}{_a\mu_g} \[1em]g​μa​=a​μg​1​So, substituting we get,gμa=132=23_{\text{g}}\text{μ}_{\text{a}} = \dfrac{1}{\dfrac{3}{2}} = \dfrac{2}{3} \[1em]g​μa​=23​1​=32​Hence, gμa = 23\dfrac{2}{3}32​

Question 4

Question 2(i)(a) A body of mass 5.5 kg is dropped from the 2nd floor of a building which is at a height of 15 m. What is the force acting on it during its fall? (g = 10 ms-2)(b) What are isobars ?(c)	How many pulleys are there in a movable block and tackle system with velocity ratio 5 ?

Accepted Answer

(a) Given,mass (m) = 5.5 kgheight (h) = 15 mforce acting on it during its fall = ?Force acting on the body during its fall = weight of the body.And F = mgSubstituting we get,F = 5.5 x 10 = 55 NHence, Force acting on the body during its fall = 55 N(b) The atoms of different elements which have the same mass number A but different atomic number Z are called isobars.Example — 1123Na^{23}_{11}	ext{Na}1123​Na and 1223Mg^{23}_{12}	ext{Mg}1223​Mg are isobars.(c) The number of pulleys required in the movable block of a block and a tackle system with velocity ratio 5 will be 5.

Question 5

Question 2(iv)(a) A brass ball is hanging from a stiff cotton thread. Draw a neat labelled diagram showing the forces acting on the brass ball and cotton thread.(b)	Can the couple acting on a rigid body produce translatory motion?

Accepted Answer

(a) In the figure below, the force on the ball is the weight W acting vertically downwards and the force on the thread is the tension T acting upwards.(b) No, the couple acting on a rigid body cannot produce translatory motion, it can cause only rotatory motion as the resultant force is zero.

Question 6

Question 2(v)Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force? Explain it with an example.

Accepted Answer

Yes, it is possible.For example, when an object is moving in a circular path due to centripetal force acting on it.At any instant of time, a constant acceleration due to the centripetal force acts on the object along the radius towards the centre while the direction of displacement is tangential to the circle i.e., force and displacement are perpendicular to each other and hence, no work is being done.

Question 7

Question 2(vi)Can SONAR be used to determine the depth of a sea? Draw a well labelled diagram showing the procedure.

Accepted Answer

Yes, SONAR is used for determining the depth of the sea.To find the depth of the sea, the time interval (t) between the instant when waves are sent and the instant when waves are received, after reflection from the depth is measured.The depth d of the sea is then —d=Vt2\text{d} = \dfrac{Vt}{2} \[0.5em]d=2Vt​where, V is the speed of ultrasonic wave in water.This process is called echo depth sounding.

Question 8

Question 2(vii)Of the three connecting wires in a household circuit:(a)	Which two of the three wires are at the same potential?(b) In which of the three wires should the switch be connected?

Accepted Answer

(a) Neutral and earth wires(b) Live wire

Question 9

Question 3(i)A lens forms an erect, magnified and virtual image of an object. Name the lens used and draw a labelled ray diagram to show the image formation.

Accepted Answer

Convex lens

Question 10

Question 3(ii)(a) The given figure shows three resistors. Find the combined resistance.(b) Which part of an electrical appliance is earthed?

Accepted Answer

(a) In the circuit, there are two parts. In the first part, resistors of R1 (6 Ω) and R2 (2 Ω) are connected in parallel. If the equivalent resistance of this part is Rp then1RP=16+121RP=1+361RP=461RP=23⇒RP=32=1.5Ω\dfrac{1}{{R}_P} = \dfrac{1}{6} + \dfrac{1}{2} \[0.5em] \dfrac{1}{{R}_P} = \dfrac{1 + 3}{6} \[0.5em] \dfrac{1}{{R}_P} = \dfrac{4}{6} \[0.5em] \dfrac{1}{{R}_P} = \dfrac{2}{3} \[0.5em] \Rightarrow {R}_P = \dfrac{3}{2} = 1.5 ΩRP​1​=61​+21​RP​1​=61+3​RP​1​=64​RP​1​=32​⇒RP​=23​=1.5ΩHence, RP = 1.5 ΩIn the second part, RP (1.5 Ω) and R3 (6 Ω) are connected in series. If the equivalent resistance of this part is Rs thenRs = (1.5 + 6) Ω = 7.5 ΩHence, combined resistance = 7.5 Ω(b) The earth wire is connected to the metallic body of the appliance.

Question 11

Question 3(iii)The diagram below shows a coil connected to the centre zero galvanometer.What polarity is induced at point X when(a) the magnet is moved towards the coil(b) the magnet is moved away from the coil?

Accepted Answer

(a) South poleReason — A moving south pole of a magnet near a coil induces a current. The induced current creates a magnetic field opposing the change, following Lenz's law. This leads to the coil's side facing the south pole developing a south pole as well.(b) North poleReason — Moving a magnet's south pole away from a coil induces a current. The induced current creates a magnetic field opposing the change, following Lenz's law. This leads to the coil's side facing the south pole developing a north pole.

Question 12

Question 3(iv)Some common salt is added to ice. It is found that temperature of mixture drops below 0°C. Explain the fact and where is this used?

Accepted Answer

When common salt is added to ice, it takes heat from ice and mixes with the water droplets present on the outer layer of ice. Hence, it absorbs heat from ice and temperature of the mixture falls below 0°C.

Question 13

Question 3(v)Suggest one effective process for the safe disposal of nuclear waste.

Accepted Answer

The nuclear waste obtained from laboratories, hospitals, scientific establishments or power plants must be first kept in thick casks and then they must be buried in the specially constructed deep underground stores. These underground stores must be far away from the populated areas.

Question 14

Question 4(i)A boy uses monochromatic green light to find the refractive index. When a ray of monochromatic green light enters in a liquid medium from air medium as shown in the figure given below, the angle 1 is 45° and angle 2 is 30°.(a)	Determine the refractive index of the liquid.(b)	Represent in the diagram showing the path of the ray after it strikes the mirror and re-enters air. Mark in the diagram wherever necessary.(c)	Draw the diagram again, if plane mirror becomes normal to the refracted ray inside the liquid. Name the principle used.

Accepted Answer

(a) Refractive index of the liquid is given by Snell's law and is shown as below,airμliquid = sin isin r\dfrac{\text{sin i}}{\text{sin r}}sin rsin i​airμliquid = sin 45°sin 30°\dfrac{\text{sin 45}\degree}{\text{sin 30}\degree}sin 30°sin 45°​=1212=22=2=1.414= \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{2}} \[1em] = \dfrac{2}{\sqrt{2}} \[1em] = \sqrt{2} \[1em] = 1.414 \[1em]=21​2​1​​=2​2​=2​=1.414Hence, the refractive index of the lens is 1.4(b) Below labelled diagram shows the path of the ray after it strikes the mirror and re-enters in air:(c) Below labelled diagram shows the path of the ray when it strikes the plane mirror normally inside the liquid:The principle of reversibility is used.

Question 15

Question 4(ii)(a) Mention one difference between reflection of light from a plane mirror and total internal reflection of light from a prism(b) When an empty test tube is placed in a beaker of water and viewed from above, its surface appears as a mirror. Explain.

Accepted Answer

(a) Difference between reflection of light from a plane mirror and total internal reflection of light from a prism	are:Plane mirrorPrismIn reflection of light from a plane mirror, only a part of the light is reflected while rest is refracted and absorbed. So, the reflection is partial.In total internal reflection of light from a prism, the entire incident light is reflected back into the denser medium. So there is no loss of light energy.(b) Total internal reflection occurs when light traveling through a medium with a higher refractive index (such as water) encounters a boundary with a medium of lower refractive index (such as air) at an angle greater than the critical angle. In this case, the water acts as a more optically dense medium than air.As a result, the light reflects completely back into the water instead of passing into the air above the test tube. This reflection creates a mirror-like effect on the surface of the water inside the test tube when viewed from above, giving the appearance of a mirror.

Question 16

Question 4(iii)If a light of a single colour is passed through a liquid having a piece of glass suspended in it, so on changing the temperature of the liquid, at a particular temperature, the glass piece is not seen.(a)	At what situation, the glass piece will not be seen?(b)	Why the light of a single colour is used?(c)	How can you define the term refractive index of a medium? State whether it can be less than 1 or not.

Accepted Answer

(a) The glass piece is not seen when the refractive index of the liquid becomes equal to the refractive index of glass.(b) The light of a single colour is used as the refractive index of a medium (glass or liquid) is different for the light of different colours.(c) The refractive index of a medium is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in that medium.μ=Speed of light in vacuum or air(c)Speed of light in that medium (V)μ = \dfrac{	ext{Speed of light in vacuum or air(c)}}{	ext{Speed of light in that medium (V)}}μ=Speed of light in that medium (V)Speed of light in vacuum or air(c)​The refractive index of a transparent medium is always greater than 1 (it cannot be less than 1) because the speed of light in any medium is always less than that in vacuum ( i.e. V < c ) and μ = 1 for air or vacuum.

Question 17

Question 5(i)(a)	Define power of a lens.(b)	A child is using a spectacle with power of -2.5D. What is meant by the negative sign?(c)	Find the focal length of the lens used.

Accepted Answer

(a) The deviation of the incident light rays produced by a lens on refraction through it is a measure of its power. It can be defined as a reciprocal of focal length:Power=1focal length\text{Power} = \dfrac{1}{\text{focal length}}Power=focal length1​(b) As the given power is negative hence, we can say that the lens used is concave in nature and the child is suffering from myopia or short sightedness (i.e., unable to see the far objects distinctly).(c) As we know,Power=1focal length\text{Power} = \dfrac{1}{\text{focal length}}Power=focal length1​Given,Power of the lens = -2.5 DSubstituting the value of power in formula we get,−2.5=1focal length⇒focal length=−12.5⇒focal length=−0.4 m⇒focal length=−40 cm-\text{2.5} = \dfrac{1}{\text{focal length}} \[0.5em] \Rightarrow \text{focal length} = -\dfrac{1}{2.5} \[0.5em] \Rightarrow \text{focal length} = -\text{0.4 m} \[0.5em] \Rightarrow \text{focal length} = -\text{40 cm} \[0.5em]−2.5=focal length1​⇒focal length=−2.51​⇒focal length=−0.4 m⇒focal length=−40 cm∴ Focal length = - 40 cm.

Question 18

Question 5(ii)The diagram given below shows a lens as combination of glass prisms of different refractive indices. Copy the diagram and answer the following questions.(a) Name the lens formed by the combination.(b) Complete the ray diagram when AB is between optical centre and 2F.

Accepted Answer

(a) The lens formed by the combination is concave in nature.(b) The completed ray diagram when AB is between optical centre and 2F is shown below:

Question 19

Question 6(i)Shyam is trying to balance a meter rod on his finger as shown in the diagram. Two weight 50 gf and 200 gf at 40 cm and 95 cm respectively and his finger at 70 cm.Find the anti-clock wise moment and clockwise moment.(b)	Find the weight of meter rod.

Accepted Answer

(a) Let W be the weight of the rod.For uniform meter rod, its weight acts at 50 cm.Clockwise moments = [200 x (95-70)] = 200 x 25 = 5000 NmAnticlockwise moments = [50 x (70 - 40)] + [W x (70 - 50)]= [50 x 30] + [W x 20]= 1500 + 20W(b) According to principle of moment :Anticlockwise moment = Clockwise moment1500 + 20W = 5000W = 5000−150020=350020\dfrac{5000 - 1500}{20} = \dfrac{3500}{20}205000−1500​=203500​ = 175 NHence, weight of meter rod = 175 NAs W = 175 N, hence, anticlockwise moment = 1500 + 20W = 1500 + (20 x 175) = 1500 + 3500 = 5000Hence, anticlockwise moment = 5000 Nm

Question 20

Question 6(ii)(a)	Draw a diagram of a pulley system of velocity ratio 4. Calculate its mechanical advantage, if its efficiency is 90%.(b)	Why does a rope walker hold a long pole in his hands? Explain.

Accepted Answer

(a) Diagram of a pulley system of velocity ratio 4 is shown below:Given,Efficiency (η) = 90%Velocity ratio = 4MA = ?We know, η=MAVR\eta = \dfrac{	ext {MA}}{	ext {VR}}η=VRMA​Substituting in the formula above we get,90100=MA4\dfrac{90}{100} = \dfrac{	ext {MA}}{	ext {4}}10090​=4MA​MA = 0.90 x 4 = 3.6Hence, MA = 3.6 N(b) The pole increases the walker's moment of inertia or resistance to rotating and prevents him from falling off the wire, allowing the walker time to correct his position and stay balanced. When he feels he is falling towards right, he shifts the pole towards left and vice versa.In addition, holding the pole low helps to keep his center of gravity low, and this further increases his ability to balance.

Solved Sample Paper 4

Section A

Section B