PHYSICS – D3 1. A uniform electric field vector E exists along horizontal direction as shown. The electric potential at A is A V . A small point change q is slowly taken from A to B along the curved path as shown. The potential energy of the charge when it is at point B is A) A q V Ex + B) A q Ex V − C) qEx D) A q V Ex − Ans: D) A B V V E x − = B A V V E x = − B B U qV = ( . ) B A U q V E x = − 2. A parallel plate capacitor of capacitance 1 C with a dielectric slab in between its plates is connected to a battery. It has a potential difference 1 V across its plates. When the dielectric slab is removed, keeping the capacitor connected to the battery, the new capacitance an d potential difference are 2 C and 2 V respectively. Then, A) 1 2 1 2 , V V C C B) 1 2 1 2 , V V C C C) 1 2 1 2 , V V C C = D) 1 2 1 2 , V V C C = Ans: C) If dielectric is removed, capacitance decreases but since capacitor is still connected to the cell, potential difference will remain the same 1 2 1 2 C C and V V = 3. A cubical Gaussian surface has side of length a = 10 cm. Electric field lines are par allel to x – axis as shown. The magnitudes of electric fields through surfaces ABCD and EFGH are 1 1 6 9 kNC and kNC − − respectively. Then the total charge enclosed by the cube is [Take 12 1 0 9 10 Fm − − = ] A) 1.35 nC B) - 1.35 nC C) 0.27 nC D) – 0.27 nC Ans: C) 0 ( )cos180 ABCD E A = 3 2 2 (6 10 )(10 10 ) ( 1) − = − 3 2 6 10 10 ( 1) − = − 60 Vm = − ( )( ) 2 3 2 0 9 10 10 10 cos 0 EFGH − = 3 2 9 10 10 90 Vm − = = Flux through other faces of cube = 0 ( ) 0 90 = 60 90 30 cube Vm = − + = ( ) 12 0 30 8.85 10 in cube in Q Q − = = 12 265.5 10 C − = 9 0.265 10 C − = 0.27 nC = 4. Electric field at a distance ‘r’ from an infinitely long uniformly charged straight conductor, having linear charge density is 1 E . Another uniformly charged conductor having same linear charge density is bent into a semicircle of radius ‘r’. The electric field at its centre is 2 E . Then A) 1 2 E E r = B) 1 2 E E = C) 1 2 E rE = D) 2 1 E rE = Ans: B) 1 0 2 E r = , 2 0 2 E r = 1 2 E E = 5. Five capacitors each of value 1 F are connected as shown in the figure. The equivalent capacitance between A and B is A) 1 F B) 2 F C) 5 F D) 3 F Ans: A) The circuit is equivalent to a balanced Wheatstone br idge. Therefore the capacitor in the middle does not get charged. 1 1 1 2 2 eq C F = + = 6. For a given electric current the drift velocity of conduction electrons in a copper wire is d V and their mobility is . When the current is increase at constant temperature A) d V remains the same, increases B) d V decreases, remains the same B) d V remains the same, decreases D) d V increases, remains the same Ans: D) d i nAev = If current is increased, drift velocity also increases. Mobility d v i l E Ae v = = i A Ae i = v i from v iR l A = = 1 e = Mobility doesn’t change. 7. Ten identical cells each emf 2 V and internal resistance 1 are connected in series with two cells wrongly connected. A resistor of 10 is connected to the combination. What is the current through the resistor ? A) 2.4 A B) 0.6 A C) 1.2 A D) 1.8 A Ans: B) 8(2) 2(2) eq = − 16 4 12 V = − = 10(1) 10 eq r = = eq eq R r R = + 10 10 20 = + = 12 3 0.6 20 5 eq eq i A R = = = = 8. The equivalent resistance between the points A and B in the following circuit is A) 5.5 B) 0.05 C) 5 D) 0.5 Ans: A) Let us assume equivalent resistance is ‘R’. Then circuit can be redrawn as 2 4 2 R R R + = + 2 8 4 2 2 R R R R + + = + 2 4 8 0 R R − − = 4 16 32 2 R + = 4 6.92 5.46 5.5 2 + = = 9. A charged particle is subjected to acceler ation in a cyclotron as shown. The charged particle undergoes increase in its speed A) Only inside 2 D B) Inside 1 2 , D D and the gaps C) Only inside 1 D D) Only in the gap between 1 2 D and D Ans: D) Electric field is responsible for increase in its speed whic h is present in between ‘Dees’ only. 10. The resistance of a carbon resistor is 4.7 5% k . The colour of the third band is A) Red B) violet C) orange D) gold Ans: A) Given Resistance 4.7 5% k = 2 47 10 5% = + Third band corresponds to number ‘2’ i.e., Red. 11. The four bands of a colour coded resistor are of the colours grey , red, gold and gold. The value of the resistance of the resistor is A) 82 10% B) 8.2 5% C) 82 5% D) 5.2 5% Ans: B) 1. Grey 2. Red 3. Gold 4. Gold Grey - 8 Red – 2 Gold - 1 10 − Gold - 5% Thus 8.2 5% 12. A wire of resistance R is connected across a cell of emf and internal resistance r. The current through the circuit is I. In time t, the work done by the battery to establish the current I is A) 2 t R B) IRt C) 2 I Rt D) It Ans: D) work done by battery to establish a current = It 13. The Curie temperatures of Cobalt and iron are 1400 K and 1000 K respectively. At T = 1600 K, the ratio of magnetic susceptibility of Cobalt to that of iron is A) 3 B) 7 5 C) 5 7 D) 1 3 Ans: A) c T Tc = − (Curie’s law) 1600 1400 200 co c c = = − 1600 1000 600 Fe c c = = − 600 3 200 co Fe c c = = 14. The torque acting on a magnetic dipole placed in uniform magnetic field is zero, when the angle between the dipole axis and the magnetic field is _________ A) 0 45 B) 0 60 C) 0 90 D) zero Ans: D) sin z MB = If 0 = , i.e. & M B are parallel to each other =0 15. The horizontal component of Earth’s magnetic field at a place is 5 3 10 T − . If the dip at that place is 0 45 , the resultant magnetic field at that place is A) 5 3 10 2 T − B) 5 3 3 10 2 T − C) 5 3 2 10 T − D) 5 3 10 T − Ans: C) 5 3 10 H B T − = 45 = cos H B B = 5 0 3 10 .cos 45 B − = 5 3 2 10 B T − = 16. A proton and an alpha - particle moving with same velocity enter a uniform magnetic field with their velocities perpendicular to the magnetic field. The ratio of radii of their circular paths is A) 1 :4 B) 4 : 1 C) 1 : 2 D) 2 : 1 Ans: C) Radius of circular path transverse of by the charged particle is given as: mv r qB = m r q p p p r m q r q m = p p m r r = e 1 2 e 2 4 1 2 p m = 17. A moving coil galvanometer is converted into an ammeter of range 0 to 5 mA. The galvanometer resistance is 90 and the shunt resistance has a value of 10 . If there are 50 divisions in the galvanometer - turned – ammeter on either sides of zero, its current sensitivity is A) 5 1 10 / A div B) 4 2 10 / A div C) 5 1 10 / div A D) 4 2 10 / div A Ans: C) 1 g s R R n = − n → Factor by which current Range has been increased. 90 10 10 1 n n = = − Current Range of Galvanometer= 5 10 mA 4 5 10 A − = Current sensitivity of Galvanometer= No of division Rangeof Galvanometer 4 50 5 10 − = 5 10 div A = 18. A positively charged particle of mass m is passed through a velocity selector. It moves horizontally rightward without deviation along the line 2 mv y qB = with a speed v. The electric field is vertically downwards and magnetic field is into the plane of the paper. Now, the electric field is switched off at t = 0. The angular momentum of the charged particle origin O at m t qB = is A) 2 3 2 mE qB B) zero C) 3 2 mE qB D) 2 3 mE qB Ans: No Correct Option Particle is moving without Deviation. qE qVB = E VB = In time of m t qB = , it completes half circular motion. At m t qB = , particle will be at ‘D’ Angular momentum about origin= r p ⊥ 4 mv mv qB = 2 2 4 m v qB = = 2 2 4 m E qB B 2 2 3 4 m E qB = 19. In series LCR circuit at resonance, the phase difference between voltage and current is A) B) 4 C) 2 D) zero Ans: D) At resonance L C X X = 0 = 20. An ideal transformer has a turns ratio of 10. When the primary is connected to 220 V, 50 Hz ac source, the power output is A) 1 10 th the power input B) equal to power input C) Zero D) 10 times the power input Ans: B ) For ideal transformer / / I P O P P P = 21. The current in a coil changes from 2A to 5A in 0.3s. The magnitude of emf induced in the coil is 1.0V. The value of self - inductance of the coil is A) 100 mH B) 0.1 mH C) 10 mH D) 1.0 mH Ans A) | | = L ∆ 𝐼 ∆ 𝑡 1 = L ( 5 − 2 ) ( 0 3 ) L= 0 3 3 = 0.1 H = 100mH 22. A metallic rod of length 1 m held along east – west direction is allowed to fall down freely. Given horizontal component of earth’s magnetic field 5 3 10 H B T − = . The emf induced in the rod at an instant t = 2s after it is released is (Take 2 10 g ms − = ) A) 3 3 10 V − B) 4 3 10 V − C) 3 6 10 V − D) 4 6 10 V − Ans D) Velocity at 2s under free fall 𝑣 = 𝑢 + 𝑎𝑡 𝑣 = 0 + 10 × 2 = 20m/s Induced emf e = Blv = 3 × 10 − 5 × 1 × 20 = 6 × 10 − 4 V 23. A square loop of side 2 cm enters a magnetic field with a constant speed of 2 cm 1 s − as shown. The front edge enters the field at t= 0s. Which of the following graph correctly depicts the induced emf in the loop? (T ake clockwise direction positive) Ans; C) From t = 0 to 1 s, emf induced = Blv = constant and current is in anticlockwise direction. 24. For a given pair of transparent media, the critical angle for which colour is maximum? (A) R ed (B) Blue (C) Violet (D) Green Ans A) According to Snell’s law Sin 𝑖 𝑐 = 1 𝑛 According to Cauchy equation 𝑛 ∝ 1 𝜆 𝜆 𝑅 > 𝜆 𝑉 𝑛 𝑅 < 𝑛 𝑉 So 𝑖 𝑐 for Red is maximum. 25. An equiconvex lens made of glass refractive index 3 2 has focal length f in air. It is completely immersed in water of refractive index 4 3 . The percentage change in the focal length is (A) 300% decrease (B) 400% decrease (C) 300% increase (D) 400% increase Ans: C) In air 1 𝑓 𝑎 = ( 𝑛 𝑙 − 1 ) ( 1 𝑅 1 − 1 𝑅 2 ) − − − ( 1 ) In medium 1 𝑓 𝑚 = ( 𝑛 𝑙 𝑛 𝑚 − 1 ) ( 1 𝑅 1 − 1 𝑅 2 ) − − − ( 2 ) ( 1 ) ( 2 ) => 𝑓 𝑚 𝑓 𝑎 = ( 𝑛 𝑙 − 1 ) ( 𝑛 𝑙 𝑛 𝑚 − 1 ) 𝑓 𝑚 𝑓 𝑎 = ( 3 2 − 1 3 2 × 3 4 − 1 ) = 1 2 × 8 1 𝑓 𝑚 𝑓 𝑎 = 4 % change in focal length = ( 𝑓 𝑚 𝑓 𝑎 − 1 ) × 100 = ( 4 − 1 ) 𝑋 3 = 300% increases 26. A point object is moving at a constant speed of 1 ms - 1 along the principal axis of a convex lens of focal length 10 cm. The speed of the image is also 1 ms - 1 , when the object is at _____ cm from the optic centre of the lens. (A) 15 (B) 20 (C) 5 (D) 10 Ans: B) Velocity of the image for convex lens 𝑑𝑣 𝑑𝑡 = ( 𝑚 2 ) 𝑑𝑢 𝑑𝑡 𝑑𝑣 𝑑𝑡 = ( 𝑓 𝑓 + 𝑢 ) 2 × 𝑑𝑢 𝑑𝑡 1 = ( 10 10 + 𝑢 ) 2 × 1 10 10 + 𝑢 = ± 1 10 10 + 𝑢 = + 1 10 10 + 𝑢 = − 1 10 =10+u 10 = − 10 − 𝑢 Not possible U = - 20cm 27. When light propagates through a given homogeneous medium, the velocities of (A) Primary wavefronts are lesser than those of secondary wavelets. (B) Primary wavefronts are greater than or equal to those of secondary wavelets. (C) Primary wavefront and wavelets are equal. (D) Primary wavefront are larger than those of secondary wavelets. Ans: C) Primary wave front and wavelets are equal because light does not change medium, it is travelling only in one homogenous medium. 28. Total impedance of a series LCR circuit varies with angular frequency of the AC source connected to its as shown in the graph. The quality factor Q of the series LCR circuit is (A) 2.5 (B) 5 (C) 1 (D) 0.4 Ans: A) 𝑄 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑓 𝑟 ∆ 𝑓 = 𝑅𝑒𝑠𝑜𝑛𝑎𝑡𝑖𝑛𝑔 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝐵𝑎𝑛𝑑 𝑤𝑖𝑑𝑡 ℎ = 500 600 − 400 = 500 200 = 2 5 29. The ratio of the magnitudes of electric field to the magnetic field of an electromagnetic wave is of the order of (A) 10 5 ms - 1 (B) 10 5 ms - 1 (C) 10 8 ms - 1 (D) 10 - 8 ms - 1 Ans: C) Speed of light = 𝐸 0 𝐵 0 = 3 8 10 / m s So, order = 10 8 𝑚 / 𝑠 30. For a point object, which of the following always produces virtual image in air? (A) Plano - convex lens (B) Convex mirror (C) Biconvex lens (D) Concave mirror Ans: B) Convex mirror always produces the virtual image for all object distance. 31. In an experiment to study photo - electric effect the observed variation of stopping potential with frequency of incident radiation is as shown in the figure. The slope and y - intercept are (A) 0 , hv v e (B) , hv h e e − (C) 0 , hv hv − (D) 0 , hv h e e − Ans: D) ℎ 𝜈 = 𝑒 𝑉 0 + ℎ 𝜈 0 𝑒 𝑉 0 = ℎ 𝜈 − ℎ 𝜈 0 𝑉 0 = ( ℎ 𝑒 ) 𝜈 − ( ℎ 𝑒 ) 𝜈 0 𝑠𝑙𝑜𝑝𝑒 = ℎ 𝑒 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = − ( ℎ 𝑒 ) 𝜈 0 32. In the Rutherford’s alpha scattering experiment, as the impact parameter increases, the scattering angle of the alpha particle (A) is always 90 0 (B) decreases (C) increases (D) r emains the same Ans: B) 𝑏 = 1 4 𝜋 ∈ 0 𝑍 𝑒 2 cot 𝜃 2 1 2 𝑚 𝑣 2 So, as impact parameter b increases, angle of scattering decreases 33. Three energy levels of the hydrogen atom and the corresponding wavelength of the emitted radiation due to different transition are as shown. Then, (A) 2 3 1 2 3 = + (B) 2 1 3 = + (C) 1 3 2 1 3 = + (D) 1 2 3 1 2 = + Ans: C) ℎ 𝑐 𝜆 2 = ℎ 𝑐 𝜆 1 + ℎ 𝑐 𝜆 3 1 𝜆 2 = 1 𝜆 1 + 1 𝜆 3 𝜆 2 = 𝜆 1 𝜆 3 𝜆 1 + 𝜆 3 34. An unpolarised light of intensity I is passed through two polaroid kept one after the other with their planes parallel to each other. The intensity of light emerging from second polaroid is 4 I . The angle between the pass axes of the polaroids is (A) 0 0 (B) 60 0 (C) 30 0 (D) 45 0 Ans: D) 𝐼 2 ( cos 𝜃 ) 2 = 𝐼 4 cos 𝜃 = 1 √ 2 𝜃 = 45° 35. In the Young’s double slit experiment, the intensity of light passing through each of the two double slits is 2 2 2 10 Wm − − . The screen slit distance is very large in comparison with slit - slit distance. The fringe width is . The distance between the central maximum and a point P on the screen is 3 x = Then the total light intensity at that point is (A) 2 2 4 10 Wm − − (B) 2 2 2 10 Wm − − (C) 2 2 16 10 Wm − − (D) 2 2 8 10 Wm − − Ans: B) 𝐼 0 = 2 × 10 − 2 𝑊 / 𝑚 2 𝐼 = 4 𝐼 0 ( cos ( 𝜋𝑥 𝛽 ) ) 2 𝑥 = 𝛽 3 𝐼 = 4 × 2 × 10 − 2 ( cos ( 𝜋𝛽 𝛽 ( 3 ) ) ) 2 = 8 × 10 − 2 ( cos ( 𝜋 ( 3 ) ) ) 2 = 8 × 10 − 2 ( 1 2 ) 2 = 2 × 10 − 2 𝑊 / 𝑚 2 36. A 60 W source emits monochromatic light of wavelength 662.5 nm. the number of photons emitted per second is (A) 20 2 10 (B) 26 5 10 (C) 29 2 10 (D) 17 5 10 Ans: A) 𝑁 = 𝑃 ℎ 𝑐 𝜆 = 𝑃𝜆 ℎ 𝑐 = 60 × 6 629 × 10 − 7 6 629 × 10 − 34 × 3 × 10 8 = 2 × 10 20 𝑝𝑒𝑟 𝑠 37. When a p - n junction diode is in forward bias, which type of charge carriers flows in the connecting wire? ( A) Ions (B) Protons (C) Holes (D) Free electrons Ans: D) In forward bias, free electrons flow in the connecting wire. 38. A full - wave rectifier with diodes D 1 and D 2 is used to rectify 50Hz alternating volage. The diode D 1 conducts ______ times in one second. (A) 25 (B) 75 (C) 50 (D) 100 Ans: C) The diode D 1 conducts 50 times in one second. 39. The truth table for the given circuit is (A) A B Y 1 1 1 1 0 0 0 1 1 0 0 1 (B) A B Y 1 1 1 1 0 1 0 1 1 0 0 1 (C) A B Y 1 1 1 1 0 0 0 1 1 0 0 0 (D) A B Y 1 1 1 1 0 1 0 1 0 0 0 1 Ans: B) 40. The energy gap of an LED is 2.4 eV. When the LED is switched ‘ON’, the momentum of the emitted photons is (A) 27 1 2.56 10 . . kg m s − − (B) 11 1 1.28 10 . . kg m s − − (C) 27 1 0.64 10 . . kg m s − − (D) 27 1 1.28 10 . . kg m s − − Ans: D) 𝐸 = 2 4 𝑒𝑉 = 2 4 × 1 6 × 10 − 19 𝐽 𝑃 = 𝐸 𝑐 = 2 4 × 1 6 × 10 − 19 3 × 10 8 = 1 28 × 10 − 27 𝑘𝑔𝑚 / 𝑠 41. In the following equation representing − decay, the number of neutrons in the nucleus X is 210 1 83 Bi X e v − → + + (A) 127 (B) 125 (C) 84 (D) 126 Ans : D) n+ p = 210 n + 83 =210 n = 210 – 83 = 127 After β - decay, number of neutrons decreases by one n = 127 - 1 =126 42. A nucleus with mass number 220 initially at rest an alpha particle. If the Q value of reaction is 5.5 MeV, calculate the value of kinetic energy of alpha particle. (A) 5.4 MeV (B) 7.4 MeV (C) 4.5 MeV (D) 6.5 MeV Ans : A) Conservation of momentum m y v y = m α v α v y = 𝑚 𝛼 𝑣 𝛼 𝑚 𝑦 Q = KEα + KE Y 5.5 = 1 2 m α v α 2 + 1 2 m y v y 2 = 1 2 m α v α 2 + 1 2 m y ( 𝑚 𝛼 𝑣 𝛼 𝑚 𝑦 ) 2 5.5 = 1 2 m α v α 2 [ 1 + 𝑚 ∝ 𝑚 𝑦 ] 5.5 MeV = K.Eα [ 1 + 4 216 ] K.Eα = 5.4 MeV 43. A radioactive sample has half - life of 3 years. The time required for the activity of the sample to reduce to 1 5 th of its initial value is about (A) 7 years (B) 15 years (C) 5 years (D) 10 years Ans : A) A = A 0 𝑒 − 𝜆𝑡 𝐴 0 5 = A 0 𝑒 − 𝜆𝑡 1 5 = 𝑒 − 𝜆𝑡 𝑒 𝜆𝑡 = 5 λ×t = 2.303 log 5 ----------- (1) We know , 𝑡 1 2 = 0 693 𝜆 λ = 0 693 3 0 693 3 × t = 2.303 log 5 t = 7 years 44. A body of mass 10 kg is kept on a horizontal surface. The coefficient of kinetic friction between the body and the surface is 0.5. A horizontal force of 60 N is applied on the body. The resulting acceleration of the body is about (A) 5 ms - 2 (B) 6 ms - 2 (C) zero (D) 1 ms - 2 Ans : D) F net = F - F k 10a = 60 - μ k N 10a = 60 – 0.5 ×10 × 10 10a = 60 – 50 a = 1m/s 2 45. A ball of mass 0.2 kg is thrown vertically down from a height of 10 m. It collides with the floor and loses 50% of its energy and then rises back to the same height. The value of its initial velocity is (A) 14 ms - 1 (B) 196 ms - 1 (C) 20 ms - 1 (D) zero A ns : A) At a height 10 m, Total energy TE 1 = 1 2 mv 2 + mgh = 1 2 mv 2 + mg ×10 After collision, Total energy, TE 2 = 1 2 × TE 1 = 1 2 × [ 1 2 mv 2 + mg ×10 ] At the highest point again at same height 10m, TE 2 = mgh 1 2 × [ 1 2 mv 2 + mg ×10 ] = mg ×10 1 4 × mv 2 = 1 2 mg ×10 v 2 = 2 × g× 10 = 2 × 9.8 × 10 v = √ 2 × √ 98 196 = = 14 m/s 46. The moment of inertia of a rigid body about an axis (A) does not depend on its shape. (B) depends on the position of axis of rotation. (C) does not depend on its size (D) does not depend on its mass. Ans : B) depends on t he position of axis of rotation 47. Seven identical discs are arranged in a planar pattern, so as to touch each other as shown in the figure. Each disc has mass ‘m’ radius R. What is the moment of inertia of system of six disc about an axis passing through the centre of central disc and norm al to plane of all discs ? (A) 100 mR 2 (B) 2 55 2 mR (C) 2 85 2 mR (D) 27 mR 2 Ans: B) I= I 1 + I 2 I 1 = moment of inertia of center disc = 𝑚 𝑅 2 2 I 2 = mom ent of inertia of six disc = 6{ 𝑚 𝑅 2 2 + 𝑚 ( 2 𝑅 ) 2 } I = 𝑚 𝑅 2 2 + 3 𝑚𝑅 2 + 24 mR 2 = 55 𝑚 𝑅 2 2 48. The true length of a wire is 3.678 cm. When the length of this wire is measured using instrument A, the length of the wire is 3.5 cm. When the length of the wire is measured using instrument B, it is found to have length 3.38 cm. Then the (A) measurement with A is more accurate while measurement with B is more precise. (B) measurement with B is more accurate and precise. (C) measurement with A is more precise while measurement with B is more accurate. (D) measurement with A is more accurate and precise. Ans : A) Measurement with A is more accurate while measurements with B is more precise 49. A body is moving along a straight line with initial velocity 0 v . Its acceleration a is constant. After t seconds, its velocity becomes v . The average velocity of the body over the given time interval is (A) 2 2 0 2 v v v at + = (B) 2 2 0 v v v at + = (C) 2 2 0 2 v v v at − = (D) 2 2 0 v v v at − = Ans: C) Average velocity = 𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝜈 2 – 𝜈 0 2 = 2as S= 𝜈 2 – 𝜈 0 2 2 𝑎 Average velocity = 2 2 0 2 v v at − 50. A particle is in uniform circular motion. Related to one complete revolution of the particle, which among the statements is incorrect ? (A) Displacement of the particle is zero. (B) Average speed of the particle is zero. (C) Average velocity of the partic le is zero. (D) Average acceleration of the particle is zero. Ans : B) Averag e speed of the particle is zero 51. 100 g of ice at 0 0 C is mixed with 100 g of water at 100 0 C. The final temperature of the mixture is [Take 5 1 3.36 10 f L J kg − = and 3 1 1 4.2 10 w S J kg k − − = ] (A) 10 0 C (B) 50 0 C (C) 1 0 C (D) 40 0 C Ans : A ) Let the final temperature is T 0 C Heat gain by ice = Heat loss by water m × L f + m × S W × ( T − 0 ) = m × S W × ( 100 − T ) 100 × 80 + 100 × 1 × T = 100 × 1 × ( 100 − T ) T = 10 0 C