# 4. DPP's solution-File

STUDY INNOVATIONSEducator en Study Innovations

DPP NO. - 65 ma2  a 2 7ma2 5.  = 12 + m    2  = 12 = 7. 1 1 1 2  2v 2 6. mgh = 2 mv2 + 2 mv2 + 2 . 5 mv2  r  1 8 = mv2 [1 + 1 + ] = 2 1. 1 18 mv2 2 9mv 2 = 5  v = 1 1 2  2v 2 13 Mx + M 3 (x + L) = 0 7. KE of the ball = mgh mv2 + 2 2 mv2   5  r  = 18 4M ML L 3 x = – 3 x = – 4 8. X = 2vt = 2v → → 1 → = 2. = 2 10 h 2. → = 0 q v  r and → = qr 3 B 4 r 3 E 4 0 r 3 → → → DPP NO. - 66  → =   (v E) = 0 50 v  E c2 1. Initially the centre of mass is at L distance from the vertical rod. 4 3. v = 100 Ve =  m( 1 )  m(0) L  Applying energy conservation  As, xcm  2     GMm  1 mv 2 =  GMm  m  m 4    v2 = R 2GM R 2  2GM R  h (R  h) 1 2GM 2GM 1  1   4 . R =    R R  h   1  h centre of mass does not move in x-direction as 4R R(R  h)  R + h = 4h  h = R/3 4. Keq. is same in all three cases. All other parameter being same, rate of energy conduction is same in all three cases. Simlarly temperature difference across any material in any wall is also same. Fx = 0. After they lie on the floor, the pin joint should be at L/4 distance from the origin shown inorder to keep the centre of mass at rest.  Finally x-displacement of the pin is L and 4 y-displacement of the pin is obviously L. Hence net displacement =  4 dT 2. H = – kA dx  dT dx dT H = kA 6, 7 & 8. Let  be the angular acceleration of rod and a be acceleration of block just after its release.  mg – T = ma (1) Now as k increases, dx becomes less (–)ve T𝑙 – mg 𝑙 = m𝑙2  .... (2) So slope becomes less (–)ve 2 3 So curve will be and a = 𝑙 (3) 3. Since  → → → = 0 q v  r , → → must be same Solving we get B 4 r 3 v  r where → = velocity of charge with respect to 5mg T = 8 and  = 3g 8𝑙 observer Let A and B are the observers Now from free body diagram of rod, let R be the reaction by hinge on rod then →  → → = → → → (vC v A )  r or → → → (v C  vB )  r R + T – mg = m a = m 1  cm 2 (v A  vB )  r  0 or → → → Solving we get (v A  vB ) || r 4. 11 = 11 (Angular momentum is conserved) As 2 decreases. 2 increases. 2 9mg R = 16 Thus T =  i.e. T decreases. Therefore the earth is completing each circle around its own axis in lesser time. K.E. = 1  2 2 DPP NO. - 67 Therefore K.E. of rotation increases. Duration of the year is dependent upon time taken to complete one revolution around the sun. 5. Using  axis theorem 1. Moment of inertia is more when mass is farther from the axis. In case of axis BC, mass distribution is closest to it and in case of axis AB mass distribution is farthest .Hence A 3 B C IBC< IAC< IAB  I > I > IH IC = ICM + my2 lx = ly = I 1 – mx2 + my2 = I 1 + m (y2–x2) = I + I + m (y2 – x2) 2 = 1.6 B P B = .8

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### 4. DPP's solution-File

• 1. DPP NO. - 65 1. Mx + 3 M (x + L) = 0 3 M 4 x = – 3 ML x = – 4 L 2. B  = 3 0 r r v q 4      and E  = 3 0 r r q 4 1     B  = 0 0 ) E v (    = 2 c E v    3. v = 100 50 Ve = R GM 2 2 1 Applying energy conservation  2 v m 2 1 R GMm   = ) h R ( GMm   v2 = h R GM 2 R GM 2    R GM 2 . 4 1 =         h R 1 R 1 GM 2  ) h R ( R h R 4 1    R + h = 4h  h = R/3 4. Keq. is same in all three cases. All other parameter being same, rate of energy conduction is same in all three cases. Simlarly temperature difference across any material in any wall is also same. 5.  = 12 ma2 + 2 2 a m         = 12 ma 7 2 = 7. 6. mgh = 2 1 mv2 + 2 1 mv2 + 2 1 . 5 2 mv2 2 r v 2       = 2 1 mv2 [1 + 1 + 5 8 ] = 2 1 mv2 5 18 = 5 mv 9 2  v = gh 9 5 7. KE of the ball = 2 1 mv2 + 2 1 5 2 mv2 2 r v 2       = 18 13 mgh 8. X = 2vt = 2v g h 2 = 2. g h 2 gh 9 5 = h 3 10 2 DPP NO. - 66 1. Initially the centre of mass is at 4 L distance from the vertical rod.             4 L m m ) 0 ( m ) ( m x , As 2 1 cm centre of mass does not move in x-direction as Fx = 0. After they lie on the floor, the pin joint should be at L/4 distance from the origin shown inorder to keep the centre of mass at rest.  Finally x-displacement of the pin is 4 L and y-displacement of the pin is obviously L. Hence net displacement = 4 L 17 16 L L 2 2  
• 2. 2. H = – kA dx dT  dx dT = kA H  Now as k increases, dx dT becomes less (–)ve So slope becomes less (–)ve So curve will be 3. Since B  = 3 0 r r v q 4      , r v    must be same where v  = velocity of charge with respect to observer Let A and B are the observers then r ) v v ( A C      = r ) v v ( B C      or 0 r ) v v ( B A       or r | | ) v v ( B A     4. 11 = 11 (Angular momentum is conserved) As 2 decreases. 2 increases. Thus T =   2 i.e. T decreases. Therefore the earth is completing each circle around its own axis in lesser time. K.E. = 2 1 2 Therefore K.E. of rotation increases. Duration of the year is dependent upon time taken to complete one revolution around the sun. 5. Using  axis theorem lx = ly 2x = 1.6 x = .8 Ma2 AB = x + M(2a)2 = 4.8 Ma2 Ans.: 4.8 Ma2 6, 7 & 8. Let  be the angular acceleration of rod and a be acceleration of block just after its release.  mg – T = ma ..... (1) T – mg 2  =  3 m 2  .... (2) and a =  .... (3) Solving we get T = 8 mg 5 and  =  8 g 3 Now from free body diagram of rod, let R be the reaction by hinge on rod R + T – mg = m acm = m 2 1  Solving we get R = 16 mg 9 DPP NO. - 67 1. Moment of inertia is more when mass is farther from the axis. In case of axis BC, mass distribution is closest to it and in case of axisAB mass distribution is farthest .Hence cm x y C 5 A B 3 IBC < IAC < IAB  I P > IB > IH IC = ICM + my2 = IB 1 – mx2 + my2 = IB 1 + m (y2 –x2 ) = IP + IB + m (y2 – x2 ) > IP + IB > IP Here IB 1 is moment of inertia of the plate about an axis perpendicular to it and passing through B.  IC > IP > IB > IH
• 3. 2. (C) Taking the origin at the centre of the plank. A B 40 kg 60 kg 40 kg 60 cm smooth x m1x1 + m2 x2 + m3x3 = 0 ( xCM = 0) (Assuming the centres of the two men are exactly at the axis shown.) 60(0) + 40(60) + 40 (–x) = 0 , x is the displacement of the block.  x = 60 cm i.e. A & B meet at the right end of the plank. 3. The slope of temperature variation is more in inner dt dQ = T . KA   T = dt dQ . KA  Slope  K 1 Larger the conductivity, smaller is the slope. 4. Let Q be the charge on the ring. The electric field at point P is  E = 0 4 1   2 / 3 2 2 ) R x ( Qx  = 0 4 1   2 / 3 2 ) R 2 ( QR The rotating charged (Q) ring is equivalent to a ring in which current  flows, such that  = R 2 Qv  The magnetic field at point P is B =  4 µ0 2 / 3 2 2 2 ) R x ( R 2    =  4 µ0 2 / 3 2 ) R 2 ( QvR  B E = v µ 1 0 0  = v c2 5. The orbital velocity, v0 = r GM Its velocity is increased by 2 times, new velocity v = r GM 2 r GM 2  = escape velocity The path is parabolic in case of escape velocity. 8. L = dt =  Fdt x = xP (because x is essentially constant during the quick blow) since, the rod starts at rest, the final values therefore satisfy L = xP.  2 1 m2  = xmv  2 x 12 v    .... (1) Another expression for v  is obtained from the given information that rod makes one revolution by the time centre reaches the dot. t = 2 and vt = d  d 2 v    .... (2) from equation 1 and 2 : d 2 x 12 2     x = d 6 2   x cannot be larger than 2  2 d 6 2      d 3   DPP NO. - 68 1. MP =        f D 1 =        5 25 1 = 6 2. Heat radiated (at temp same temp)  A  Q  4R2 and Q'  (4R2 + 2 × R2 )  2 2 R 4 R 6 Q ' Q    = 1.5 Here R2 is extra surface area of plane surface of one of the hemisphere.
• 4. 3. Megnetic moment M  = r2 i î &  B  = ĵ 4 î 3     = M  × B  = r2 (3 î – 4 î )   will be along the direction shown . Hence , the point about which the loop will be lift up will be : (3, 4) 4. Radius of Curvature = on Accelerati Normal ) velocity ( 2 = R / v ) v 2 ( 2 2 = 4R 5. By linear momentum conservation in horizontal di- rection = for (bob + string + cart) mV0 = (m + m)v v = 2 V0 By mechanical energy conservation for (bob + string + cart + earth) 2 1 mV0 2 + 0 + 0 = 2 1 (2m)v2 + mgh + 0 2 1 mV0 2 – 2 1 (2m) 4 V 2 0 = mgh Solving it, h = g 4 V2 0 . 6. When two drops of radius r each combine to form a big drop, the radius of big drop will be given by 3 R 3 4  = 3 2 r 3 4 r 3 4    or R3 = 2r3 or R = r 2 3 / 1 Now 2 r R r R V V        = 3 1 3 2 4 2   VR = 5 × 41/3 cm/s 7. If m is pole strength , then m = l M m  When the wire is bent into a semicircular arc, the separation between the two poles changes from l to 2l, where new magnetic moment of the steel wire,        M 2 2 M r 2 m ' M l l 8. (A) Real image of a real object is formed by concave mirror and convex lens. (B) Virtual image of a real object is formed by all four. (C) Real image of a virtual object may be formed by all four. (D) Virtual image of a virtual object may be formed by convex mirror and concave lens. (A) p,r (B) p,q,r,s (C) p,q,r,s (D) q,s 9. Let E1 < E2 and a current i flows through the circuit. Then the potential difference across cell of emf E1 is E1 + ir1 which is positive, hence potential difference across this cell cannot be zero. Hence statement 1 is correct. For current in the circuit to be zero, emf of both the cells should be equal. But E1  E2 . Hence statement 2 is correct but it is not a correct explanation of state- ment 1. E , r 1 1 E , r 2 2 DPP NO. - 69 3. Equal area means equal power output. A3 area pertains to highest wavelength range, thus photons with minimum range of frequency. Thus maximum number of photons are required from this segment to keep the power same. 4. Work done by kinetic friction may be positive when it acts along motion of the body. Friction on rigid body rolling on inclined plane is along upward because tendency of slipping is downwards. 5. The torque of system = Torque on loop [AFGH + BCPE + ABEF] = SB ) î ( + SB( î )+ SB k̂ (I = current, S = area of loop, B = magnetic field. =  S B k̂ = 1 × 1 × 2 k̂ = 2 k̂ units [Ans: K̂ 2 ]
• 5. 6. The work done to rotate a bar magnet from its initial position  = 1 to the final position  = 2 is given by W = M B (cos 1 – cos2 ), (i) Here 1 = 0° and 2 = 180°  W = M B (cos 1 – cos180°) = M B = [1–(–1)] = 2 M B (ii) Here 1 = 0° and 2 = 90°  W = M B (cos0° – cos 90°) = M B = [1– 0)] = M B 7. If velocity of m2 is zero then by momentum conservation m1 v = m2 v v = 1 2 m v m Now kinetic energy of m1 = 2 1 m1 v2 = 2 1 m1 2 2 1 2 v m m         = 2 1         1 2 m m m2 v2 =         1 2 m m 2 1 m2 v2 = 1 2 m m × initial Kinetic energy Kinetic energy of m1 > initial mechanical energy of system m1 Hence proved 8. C = sin–1       1 / 2 1 = 30° for i = 37 TR so,  =  – 2 (37°) = 104° i = 25, Refraction  < 2  – C i = 45°, TR so,  =  – 2        4 = 90° By applying snells law for prism : i = 90 r1 = 30 r2 = 30 e = 45  = 90 + 45 – 60 = 75° 9. The pointsAand B are at same potential, then under given conditions pointsAand B on the circuit can be connected by a conducting wire. Hence the circuit can be redrawn as shown in figure 2. E B A E R1 R2 R3 Fig-1 Therefore statement 1 is true. Statement 2 is obvi- ously false. DPP NO. - 70 1. In normal adjustment m = e 0 f f so 50 = e f 100  fe = 2 cm ( eyepiece is concave lens) and L = f0 – fe = 100 – 2 = 98 cm 2. m T = const. ln m + lnT = C 0 T dT d m m      m m d   = T dT  Now m m d   = – 1 % = 100 1  (–ve sign indicates decrease) dT = 1 (given)  T = 100 K. 3. Emissive power =  T4 = 6 × 10–8 × 1004 W/m2
• 6. 4. to reach  = 270º, it has to cross the potential energy barrier at  = 180º and to cross  = 180º angular velocity at  = 180º should be 0+ ki + Ui = kf + Uy 2 2 MR 2 3 2 1        + (–Mi AB cos 0º) = 0 + (–NiAB cos180º)  = 9 80  rad/sec. 5. Equation for linear motion mgsin – f = ma for rotary motion f. R =  . R a  f = 2 R  . a mgsin = ma + 2 2 R . 2 MR .a = 2 3 ma a = 3 sin g 2  = 3 g using S = ut + 2 1 at2 for linear motion. 1 = 0 + 2 1 . 3 g . t2 t = g 6 sec. Ans. 6. Here BH = 0.22 T ; BV = 0.38 T Now 2 V 2 H B B B   = 1928 . 0 ) 38 . 0 ( ) 22 . 0 ( 2 2   = 0.44 T 7. w.r.t. the wedge As maximum height = 125 m  block want by a height 20m over the wedge  (v sin53º)2 = 2.g.20 v2 25 16 = 400 v2 = 25 × 25 v = 25 m/sec.  block left the wedge with a relative velocity 25 m/sec. Now, time of flight = 2 + 5 = 7 sec. horizontal range w.r.t. wedge = vx × T = 25 cos 53 × 7 = 105 m. 8. For slab no deviation so  = 0 for any i for slab for light from D to R  = r – i ....(i) nd sin i = nr sin r  r = sin–1       i sin n n r d  = sin–1       i sin n n r d – i it is non-linear function and graph is After i > C T.R. will occur and graph is straight line for D to R
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