Sec: XII-IIT Date: 10-03-2021

ASSIGNMENT-06

KEY Physics : Current Electricity

1 A 2 D 3 C 4 C 5 B

6 B 7 A 8 A 9 C 10 B

11 D 12 C 13 A 14 C 15 C

16 C 17 A 18 B 19 C 20 A

21 C 22 D 23 D 24 C 25 A

26 A 27 C 28 A 29 C 30 C

31 2 32 4 33 4 34 7 35 6

36 53.57 37 4.8 38 431.4 39 5 40 0.408

Physics : Current Electricity

1. (A) The Current in 2 resistor is given as

2

12

i A

Potential difference across 2 resistors is given as

10 2 20q C

2. (D ) In the current carrying wire due to external electric field when free electrons of a metal body drifts their

rate of collision increases with neighbouring electrons compared to the case when there is no electric

current due to which heat is dissipated and temperature of metal wire rises.

3. (C) At any time inside a metal body velocities of free electrons are randomly distributed in different

directions average of which is zero

4. (B) in above circuit brightness will be proportional to the current through the bulb as power dissipated is

2i R . In this case maximum current will pass through A

5. (B) Drift speed of free electrons in a conductor carrying a current i and cross sectional area A is given as

2d

i iv

neA ne r

dv2

i

r

Thus drift velocity drift to half.

6. (B) At t=0, when capacitor is undercharged it behaves like short circuit. In this case 6 and 3 will act as in

parallel combination thus total resistance across battery is

eR 1 2 3q

Current through battery and 1 resistance is 12

3i = 4A

7. (A) 1 1 2 2CV C V

1 2

2 1

C V

C V

5 5

3 3

i

i

8. (A) the charge of the capacitor becomes 1

eof initial charge in time, .

CR

0 1K A d

d A

0K

9. (C) 20

2080 80

RR

10. (B) 2 2

1

1

220484

100

VR

P and

2 2

2

2

220806.7

60

VR

P

Current, 220

0.17484 806.7

i A

Now, 2 2

1 1 0.17 484 14P i R W

11. (D) 1 21.5 1 , 1.5 2 ,........E V E V

Now Current 1 2

1 2

....

.....

n

n

E E EI

r r r

1.5 1 2 .........

1 2 ......

N

N

=1.5 A

12. (C) The equivalent resistance across the cell is , 1.5R .

Current, 6

41.5

i A

13. (A) by short circuiting the battery the effective circuit across C is shown in figure below for calculation of

Thevenin’s resistance

e

3R 1.5

2q

RR

eR 1.5eq qC CR

14. (C) When K is open equivalent resistance of the circuit is given as

e 1R 3 / 2q R

1

2/ 3 / 2

3

Ei E R

R

When K is closed equivalent resistance of the circuit is given as

e 2

2 4R 2

2 3q

R RR

R R

2

3/ 4 / 3

4

Ei E R

R

1

2

8

9

i

i

15. (C) if the unknown resistances are P and Q, then for first case we have at null point 20 Cm

204 ............( )

100 20

Por Q P i

Q

P Q

In the second case we have null point at 40 Cm

15 40

100 40

P

Q

15 2.............( )

3

Pii

Q

Solving above equation we get,

9P

16. (C) We use for 3 V battery

E=ir=0

3 15

3 1 01 2 R

Solving this equation we get,

3R

17. (A) In parallel current distributes in ration of resistance so we use

G

S

I S

I G

S

G

IG S

I

50 20

1220

G

18G

18. (B) In steady state current in the branch containing the capacitor is zero so current flows in outer loop and its

gives the potential difference across capacitor which is same as potential difference across 2R given as

2

2

2ER

VR r

2q CV C

19. (C) The given time can be considered as the half life time of the exponentially decreasing equation which is

given as

1/2 2 2Ct In In RC

2

tR

In C

2 24

2 0.5

In sR

In F

Thus resistance of ammeter equal to 2

CHEMISTRY : Electrochemistry

1 B 2 B 3 A 4 D 5 C

6 D 7 C 8 D 9 A 10 D

11 B 12 B 13 D 14 D 15 C

16 B 17 B 18 A 19 D 20 A

21 C 22 D 23 A 24 C 25 B

26 D 27 A 28 A 29 C 30 D

31 2 32 8 33 7 34 5 35 3

36 1.159 37 193 38 28.95 39 0.02486 40 40

Chemistry

4 (D)

𝐸cell° = 𝐸𝑂𝑃Sn

° +𝐸𝑅𝑃Fe

° = 0.14 + (⎯0.44)

= −0.30 V

11. Cell reaction is, Zn + Fe2+ Zn2+ + Fe

Using, 2

3

0.059 10E E log

2 10

(n = 2, [Zn2+] = 10–2, [Fe2+] = 10–3)

Since E = 0.2905

0.2905 = E° - 0.029

or E° + 0.2905 + 0.0295 = 0.3200 V

Again eq

0.059E logK

2

0.32

0.0295eq eq

0.0590.32 log K or K 10

2

12. Case I: 1C 0.2 M, R 50 , k 1.4S m

1 1

. 1.4 .50

l lk or

A R A

11.4 50 70l

mA

Case II: 170 , 0.5 , 280l

m C M RA

,

1 1 1 1

70280

l lR or

A R A

110.25k S m

Now, 1000

m kC

If molarity is in mol L–1, then

m (S m2 mol–1)

1

3 11000

k S m

L m Molarity mol L

1

3 1

0.25

1000 0.5

S m

L m mol L

4 2 15 10 S m mol

13. (A)

More negative is the standard reduction potential, greater is the tendency to lose electrons and hence,

greater is the reactivity

14. Higher the acidity, higher will be the tendency to release protons and hence lighter will be the electrical

conductivity. Difluoroacetic acid will be strongest acid due to electron withdrawing effect of two fluorine

atoms so as it will show maximum electrical conductivity.

15. (C) Metal having more EoOP replaces the other form its salt solutions.

16. (B)

Weight of Ag required = 80 5 10-3 1.05(wt. V d) = 0.42 g

. .

96500

E i tw

108 3

0.4296500

t

125sec.t

17. (B)

Conceptual

18. (A)

23 (A)

In MnO4− the oxidation number of Mn is + 7.

+7 +2

∴ Mn + 5𝑒− ⟶ Mn

In the reaction, 5 electrons are involved hence 5 Faraday will be needed for the reduction of 1 mole of MnO4

−.

Therefore, for 0.5 mole of MnO4−, number of Faradays required = 2.5 F

24. (C)

Given fact is : 𝑋 is above 𝑍 and 𝑌 in electrochemical series and thus, 𝑍 cannot displace 𝑋 from its salt.

25 (B)

NaCl + KNO3 ⟶ NaNO3 + KCl

KCl – NaCl

Molar conductivities 152 128

Difference = 24 S cm2 mol−1

Molar conductivities of KNO3 = 111

Molar conductivities of NaNO3 = 111 – 24

= 87 S cm2 mol−1 26 (D)

Given, mass of nitrobenzene = 12.3 g

Molecular mass of nitrobenzene = 123 g

∴ Moles = 12.3

123 = 0.1

∴ H2 required for reduction of 0.1 mole nitrobenzene

= 0.1 × 3 = 0.3 mol of hydrogen

∴ Amount of charge required to liberate 0.3 mol of hydrogen = 2 × 96500 × 0.3 (∵ H2 ⟶ 2H+)

= 57900 C

27 (A)

At cathode 2H+ + 2𝑒− → H2

At anode 2OH− → H2O +1

2O2 + 2𝑒−

28 (A)

2Br−(𝑎𝑞) + Cl2 ⟶ 2Cl−(𝑎𝑞) + Br2

Bromine is relased by chlorine as more reactive halogen displaces less reactive halogen from their

salt solutions.

29 (c)

The equivalent conductance of strong electrolyte is increased by dilution, because its value is equal to

the multiple of 𝐾𝑣 and the volume of solution. By making dilution the volume of solution increases

which also increase the value of equivalent conductivity.

MATHS : Vector & 3D

1 C 2 A 3 B 4 D 5 B

6 C 7 B 8 D 9 A 10 C

11 C 12 B 13 C 14 C 15 B

16 D 17 B 18 A 19 D 20 C

21 A 22 D 23 B 24 B 25 C

26 A 27 B 28 A 29 C 30 C

31 5 32 0 33 5 34 7 35 1

36 1.5 37 4.5 38 14 39 18 40 0.71

MATHEMATICS

1. A vector bisecting the angle between a and b is

|b|

b

|a|

a ; in this case

6

kj2–i

6

k–ji2

(i.e.) 6

j–i3 or

6

k2–j3i

A vector of magnitude 3 along these vectors is 10

)j–i3(3or

14

)k2–j3i(3

Now, 14

3 (i + 3j – 2k). (i – 2j + k) is negative and hence

14

3 (i + 3j – 2k) makes an obtuse angle with b

2. If the incentre I be chosen as the origin and a , b , c be the position vectors of A,B,C then the position vector

of I =

cba

But position vector of I is zero, since it is chosen as the origin.

cba =

0 a + b + c =

0

3.

AB+

AC = 2

AD

AD = 2

1 {(–3

i + 4

k ) + (5

i – 2

j + 4

k )} = (

i –

j + 4

k )

Length of AD = 1611 = 18

4. If ^i ,

^j ,

^k are the unit vectors in the oxyz system and

^i 1,

^j 1,

^k 1 are the unit vectors in the system o x y z

obtained after rotation, then

^i 1 =

2

1 ^i +

2

1

^j

^j 1 =

2

1 ^i +

2

1 ^j and

^k 1 =

^k

= 22^i 1 + 23

^j 1 + 4

^k 1

= 22

^^j

2

1i

2

1+ 23

^^j

2

1i

2

1– + 4

^k = –

^i + 5

^j + 4

^k

So that the components in the oxyz system are – 1, 5, 4.

5.

PQ ×

RS –

QR ×

PS +

RP ×

QS

= (

PS +

SQ ) ×

RS –

QR ×

PS + (

RS +

SP ) ×

QS

=

PS × (

RS +

QS +

QR ) –

SQ ×

SR –

SQ ×

SR

=

PS ×

0 – 2

SQ ×

SR = 4 × area of QRS

6. Let p = x

i + y

j + z

k

p ×

i = (x

i + y

j + z

k ) ×

i = z

j – y

k |

p ×

i |2 = z2 + y2

2

3 |

p ×

i |2 =

2

3(2x2 + 2y2 + 2z2) = 3(x2 + y2 + z2) = 3 |

p |2 = 3

p 2

7.

a ×

b = 2

a ×

c

a × (

b – 2

c ) = 0

b – 2

c =

a

Taking modulus on both sides and then squaring we get, |

b – 2

c |2 = 2 |

a |2

|

b |2 + 4|

c |2 – 4|

b | |

c | cos = 2|

a |2

16 + 4 – 4 . 4 . 1. 4

1 = 2.1

2 = 16 = ± 4

8. Scalar Triple product =

112

111

111

]cba[

9. The vectors reciprocal to a ,

b ,

c are given by

p =

)cba(

cb

,

)cba(

ac

,

)cba(

ba

= so that

a .

p = 1,

a .

q =

a .

r = 0, b . q = 1,

c . q =

a .

q = 0, r.c = 1, p.c = r.c = 0

This gives (a + m

b + n

c ). (

p + m

q + n

r )

= 2 + m2 + n2.

10. Let c

= ( a

+ b

) + ( a

+ b

)

taking dot product with a

, we get

2 = [4 + 2] = 3

1

squaring 4 = 2[4 + 4 + 4] = 2. 12

2 = 9

2

taking dot product with ( a

× b

)

[( a

× b

). c

] = (12)

= 3

2× 12 = 4 2 Q.

11. Let c

= ( a

+ b

) + ( a

+ b

)

taking dot product with a

, we get

2 = [4 + 2] = 3

1

squaring 4 = 2 [4 + 4 + 4] = 2. 12

2 = 9

2

taking dot product with ( a

× b

)

[ ( a

× b

). c

] = (12)

= 3

2× 12 = 4 2 Volume of tetrahedron is equal to 1/6. 4 2 =

3

22

12. Any such vector is [( i – j – k ) × {( i + j + k ) × (– i + j + k )}

13. E d . [a × (

b × c × d )]

= d [a × (

b × d ) c – (

b × d ) d

= d ×[(b × d ) (

a × c ) – (

a × d ) (

b × c )]

E (b ×

a ) [ d

a c ]– (

b × c )

[ d ×(a × d )]/2

= (b × d ) [ d

a c ]

Putting values, we get E = 98

14.

x ,

y and

x ×

y are non-coplanar vectors20a = 15b = 12c

3

a=

4

b =

5

c c2 = a2 + b2 .Hence right angled triangle.

15.

r ×

a =

b

a × (

r ×

a ) =

a ×

b

(

a .

a )

r – (

a .

r )

a =

a ×

b

r = (

a .

r )

a + (

a ×

b ) …(i)

Now, |

r ×

a |2 = |

b |2 = 2

or, 2 = (

r ×

a ) . (

r ×

a )

=

a.ar.a

r.ar.r

r .

a = ± 1. …(ii)

r = 3

1(

a ×

b ±

a ) from (i) & (ii)

16. Let c

= a

+ b

+ ( a

× b

), then

a

. c

= = cos

Similarly = cos

also, | c

|2 = 2 + 2 +2 | a

× b

|2

1 = 22 + 2 [ 222 )b.a(|b||a|

] = 22 + 2 [ = ]

v2 = 1 –22 = 1 – 2 cos2 = –cos2

Now 2 cos 24

4

3

17. |a|

= |b|

= |c|

= 1

angle between b

and c

= angle between a

and c

c.b

= c.a

= cos ( |a|

= |b|

= |c|

= 1)

also ba

b.a

= 0

so )ba(nbmac

c.a

= = cos, c.b

= cos = m

Now | c

| = 1 c.c

= 1 = 22 + n2| ba

|2

m = = cos

1 = 22 + n2 n2 = 1 – 22

18. a × (

b ×

c ) = (

a .

c )

b – (

a .

b )

c

= – 7b and this is orthogonal to

a .

19. ( a

× b

) × ( c

× r

) = A

× ( c

× r

) where A

= a

× b

= ( a

× b

) × B

where B

= c

× r

Now A

× ( c

× r

) = ( A

. r

) c

– ( A

. c

) r

= [ a

b

r

] c

– [ a

b

c

] r

similarly ( a

× b

) × B

= [ a

c

r

] b

– [ b

c

r

] a

so [ a

b

r

] c

–[ a

b

c

] r

=[ a

c

r

] b

–[ b

c

r

] a

[ a

b

r

] c

–[ a

b

c

] r

= – [ c

a

r

] b

–[ b

c

r

] a

[ a

b

r

] c

+ [ c

a

r

] b

+ [ b

c

r

] a

= [ a

b

c

] r

but [ a

b

c

] r

= [ b

c

a

] r

= [ c

a

b

] r

20.

D C

B A a

b

–

–

AC = a + b

| AC | = | a + b |

| AC |2 = |a|2 + |b|2 + 2 a . b

= (3 – )2 + ( + 3)2 + 2(3 – ). ( + 3 )

= 162 + 42 +16 .

= 16. 4 + 4. 4 + 16|| || cos 60° = 112

|AC| = 4 7

21. ab)]ba2()ab[()ba(

)ab)}(ba2).(ba{(

).ba{(

)ab()ba2)}(ab(

ab)ab)(1b.a2(

either 0ab

or 1b.a1

either a–b

or 0b.a

either = or = 2

22. The given lines intersect, if the shortest distance between the lines is zero.

We know that the shortest distance between the lines r = a1 + ( 1b

) and r = a2 + b2 is

|bb|

|bb).aa(|

21

2121

So the shortest distance between the given lines is zero if

(i – j – (2i – j) . (2i + k) × (i + j – k) = 0

L.H.S. =

111

102

001

= 1 0

Hence the given lines do not intersect.

23. dr’s of diagonal through the origin are (1, 1, 1)

O

z

x

y

dr's of edge, parallel to x-axis

and in the plane z = 0, are (1, 0, 0)

Hence d.c. of the line of shortest distance are

2

1–,

2

1,0

Projection of line segment joining (1, 1, 1) and (1, 1, 0) on the line of shortest distance

= – 2

1

24. Angle between the diagonal of a cube and a diagonal of one face = cos–1

3

2

25. Mid point of BC is

2

2,4,

2

1 .

DR's of median through A are -

2

1 – 2, 4 – 3,

2

2 – 5i.e.

2

5, 1,

8

2

The medium is equally inclined to axes,

D.R's must be equal 2

5 = 1 =

8

2

= 7 and µ = 10

26. For n = – ( + m), the second relation gives

a2 + bm2 + c( + m)2 = 0 or (a + c) 2 + 2cm + (b + c) m2 = 0.

For parallel lines, the two roots must be equal

4c2 – 4(b + c) (a + c) = 0 ab + bc + ca = 0

27. For = –m +

n, the second relation gives

mn + n

nm–m

nm– = 0

or n2 + mn – m2 = 0 m

n

m

n2

2

– = 0.

If 1

1

m

n,

2

2

m

nare its roots then

21

21

mm

nn = –

212121 nn

1–

mm

1–

.Since the lines are perpendicular –1 –1 + = 0 = 2

28. The given lines intersects each other if

12

11

544312

= 0

= 0, –1

29. The coordinates of a point on the given line are (2r + 1, –3r – 1, r)

The distance of this point from the point (1, –1, 0) is given to be 4 14 .

(2r)2 + (–3r)2 + (r)2 = (4 14 )2

14r2 = 16 × 14 r = ± 4

So the coordinates of the required point are

(9, – 13, 4) or (–7, 11, –4).

30. Let OP = r

1 = r sin cos … (1)

2 = r sin sin … (2)

3 = r cos … (3)

square and adding r2 = 14 r = ± 14 … (4)

Using (1), (2) (3) and (4), we get (C) as answer.

Integer Type Questions:

31. The required line through the point P(2, 1, 1) in the plane 2x + y –5z = 0 and of greatest slope is

perpendicular to line of intersection of the planes

2x + y – 5z = 0 and 4x – 3y + 7z = 0

Let dr's. of line of intersection are a, b, c then

2a + b –5c = 0 and 4a –3b + 7c = 0

as dr's of line (a, b, c) is perpendicular to dr's. of normal to both the planes.

4

a=

17

b=

5

c

Now let the dr's of required line be , m, n

then

2x =

m

1y =

n

1z where 2 + m – 5n = 0

4 + 17m + 5n = 0

So 3

=

1

m

=

1

n

3

2x =

1

1y

=

1

1z = k

Let any point on this line (2 + 3k, 1 – k, 1 + k)

also on first plane

4 (2 + 3k) – 3(1 – k) + 7 (1 + k) = 0

8 + 12k – 3 + 3k + 7 + 7k = 0

k = 11

6.so point

11

61,

11

61,

11

182

So x + y + z = 4 + 11

18.greatest integral value = 5

32. Planes ax + by = 0 and z = 0

equation of any plane through the line of intersection of these planes is ax + by + kz = 0;

k R.Hence dc's of normal to this plane

222 kba

a

,222 kba

b

,222 kba

k

dc's of normal to plane 1 are22 ba

a

, 22 ba

b

, 0

so angle cos = 22222

22

bakba

ba

=

222

22

kba

ba

k2 cos2 = 2 (1 – cos2 ) + b2 (1– cos2 )

k = ± 22 ba tan

So equation of plane

ax + by ± z 22 ba tan = 0.Hence c = 0

33. Plane contains the line

So (a) plane is r to i2 – j3 + k5 so 2a + 3b + 5c = 0 ……..(i)

(b) plane contain the line i – 2 j so a + 2b + 4 = 0……..(ii)

(c) plane is r to x – y + z + 2 = 0 so a + b + c = 0……..(iii)

on solving (i) + (ii) + (iii)

a = 2, b = – 3, c = 1

34. Let reflection of P(1, 0, 0) in the line8

10z

3

1y

2

1x

be ()

then

2,

2,

2

1lies on the line.

& (–1) kji is perpendicular to k8j3i2

8

102

3

12

2

12

1

=

and 2( – 1) – 3() + (8) = 0

= 5, = – 8, = – 4

35. Given planes are

x – cy – bz = 0 ... (i);cx – y + az = 0 ... (ii); bx + ay – z = 0 ... (iii)

Equation of plane passing through the line of intersection of plane (i) and (ii) may be taken as

(x – cy –bz) + (cx – y + az) = 0 (1 + c) x – y (c + ) + z (a – b) = 0 ....(iv)

If plane (iii) and (iv) are same thenb

c1 =

a

)c( =

1

ba

= –bac

bca

= –

2a1

cab

a – a3 + bc – a2 bc = a2 bc + ac2 + ab2 + bc

a2 + b2 + c2 + 2abc = 1

Numerical Based Questions:

36. Area of quadrilateral OABC = OAC + ABC

= 2

1ACOA +

2

1BCAB

= 2

1| a

× ( b

– a

) | +2

1|(2 a

+ 10 b

– a

) × ( b

– 2 a

–10 b

)|

= 2

1| a

× b

| +2

1|( a

+ 10 b

) × (2 a

+ 9 b

)|

= 2

1| a

× b

| + 2

11| a

× b

| = 6 | a

× b

|

| a

× b

| = m

= 2m 6 | a

× b

| = 2 | a

× b

| = 3.

37. Equation of plane is a

z

a

y

a

x = 1

It passes through the point (1,1,1).

So a

1

a

1

a

1 = 1

a = 3

hence A(3, 0, 0), B(0, 3, 0), C(0, 0, 3)

volume of tetrahedron = 6

1

2

9

300

030

003

= 2

9 2 = 9 = 4.5

38. A(1, 1, 1), B(2, 3, 5), C(–1, 0, 2) directions ratios of AB are (1, 2, 4) direction ratio of AC are

( – 2, – 1, 1) direction ratios of normal to plane ABC are (2, – 3, 1)

Equation of the plane ABC is 2x – 3y + z = 0

Let the equation of the required plane be 2x – 3y + z = k, then 194

k

= 2

k = ± 142 Equation of the required plane is 2x – 3y + z + 142 = 0

39. Given planes are:

x –z = 0, x –2y + z = 0, x + y + z = 0

Let the point whose locus is required be P(). According to question

2

|| 2+

6

|2| 2+

3

|| 2= 36

or 3(2 + 2–2) + 2 + 42 + 2 –4 –4 + 2 +2(2 +2 +2+ 2 + 2 + 2) = 36 × 6

or 62 + 62 + 62 = 36 × 6 or 2 + 2 + 2 = 36

Hence, the required equation of locus is x2 + y2 + z2 = 36

40. dr's of AL a, – a, – a

dr's of AP a, a, a

N M

P(a, a, a)

(a, 0, 0) X

C

O A

L (0, a, a)

B

y

between AL and AP

cos = 222222 aa0)a()a(a

)a(a)a(a0.a

= – 3

2 tan =

2

1