SOLUTION & ANSWER FOR ISAT-2010 – PAPER - Ι VERSION – A [PHYSICS, CHEMISTRY & MATHEMATICS] →
1.
A block A with initial velocity v 0 strikes a spring
4.
An infinitely large surface of uniform charge density σ has a disc of radius R cut out -------------
tied to a block -----→
→
Ans :
→
Ans : v 1 + v 2 = v 0 Sol:
Sol:
At maximum compression, no relative velocity. v1 = v2 = v’ for conservation of momentum, 2 mv’ = mv ⇒ v’ = →
→
a 2
R + a2
For a disc of charge density σ and radius R the electric field at a distance a σ a = 1− 2ε 0 2 2 1/ 2 R +a
(
v 2
→
)
Hence in the given problem, subtract from σ the field of the infinite sheet the 2ε 0
v1+ v 2 = v 0
2.
σ 2ε 0
A particle moves in the first quadrant of the x-y plane under the action of a -----------
above value.
σ 2ε 0
Hence E =
Ans : vx = u(1 + t / t0) Sol:
x=
5.
dx = (option given) dt
vx =
∫ (option given) dt
a 2
R + a2
All the five capacitors shown in the figure have the same capacitance C. The battery ------------Ans : Zero
This gives x = 0 at t = 0 and at no other value of t only for option (c).
Sol: P
3.
Q
V
The figure, on the right, shows a jar filled with two liquids of densities ρ and -------------------R
Ans :
S •C
A•
F
T • B O
Points A, B and C are at the same potential. Hence no charge on R, S and T.
h
Sol: Higher density liquid is lower. As the cylinder is getting immersed, the buoyancy force increases from zero linear till it is fully immersed. Once fully immersed, until it reaches the interface buoyancy force is constant. When it crosses the interface and moves into the higher density liquid, again, the buoyancy force increases linearly.
6.
The figure shows a wire mesh of infinite extent, such that the resistance --------------------Ans :
1 A 2
Sol:
i/4 i/4
P i/4
Once fully immersed in the second liquid the buoyancy force is constant.
i/4
i/4
Q i/4
i/4 i/4
P
Q i/4 + i/4
7.
Sol:
i 1 = = A 2 2
In a Young’s double slit experiment, the separation between the slits --------------------Ans : 0.25 mm Sol:
Position of the first dark fringe is given by β λD = 2 2d The least distance of the dark fringe from the centre is for the smaller wavelength.
∴
8.
500 × 10
−6
2 × 10
−3
×1
Total surface energy = T. area 2 = T × 2 × 4πr 2 = 8πr T
11. In a column of air at a given temperature T, the density ρ is found to vary with the -----------------k T Ans : B mg Sol:
Probability of finding a molecule at height z is ∞
P(z) ∝
0
= 0.25 mm
2 1 = kBT ∝ mg mg k T B
t and the magnetic field
12. The mass density in a nucleus near its center, in 3 units of kg/m , is in the range ----------------
c .ε 0 E 02 kˆ Ans : 2
Ans : 10
9.
− mgz dz kBT
∫ zρ0e 2
→ → The electric field E r , -----------------
Sol:
2
The rate of energy flow across a unit area is the time average over a cycle of the pointing vector. 1 ∴ S = ε 0 cE 02 2
Sol:
A transverse wave travels on a taut string stretched along the x-axis. The linear mass -------------Ans : A(X)
15
to 10
20
Density at the centre ρ0 ρ0 ∝ = R 1 + e −2 − 1+ e a 0.17 3 = nucleons / fm 1.135 ≅
0.15 × 10 −27 10
− 45
~ 1.5 × 10
17
kg m−
3
214
13. For the nucleus Te, the value of r for which the nucleon density falls to half its ---------------Ans : 6 to 7 fm 0
Sol:
x
ρ0
Sol:
r a −12 1+ e
Intensity of the wave down the string of linear density is proportional to A ×µ×v∝ 2
A 2µ µ
∝A
2
µ
Since energy independent of x, 1 2 A ∝ µ
=
r −12
∴ 1 + e a
(
ρ0
2 1 + e −12
)
= 2(1 + e− ) 12
⇒ e− [e − 2] = 1 r/a 12 (e − 2) = e 12
∴
r/a
r ≅ 12 a 12
10. A spherical soap bubble of radius R is blown from a tiny drop of a soap --------------Ans : 8πTr
2
(neglecting of no comparison to e ) ∴ r = 12 a = 0.55 × 12 fm = 6.6 fm
28
14. The plots of p(r) versus r for Si and another nucleus X are shown -----------------Ans :
142
Sol:
The electron current occurs above the threshold voltage. Initially the current rises steeply and then slowly attains saturation.
Ba 18. In the following statements, A Ideal gases are liquids only
ρ0 ρ0 Sol: = ( r / a −10 ) 1+ e 2 1 + e −10 ( ) 10 ∴ 1 + e r / a −10 = 2(1 + e− )
(
)
Ans : B, C, D
e− (e − 2) = 1 r/a 10 ∴ (e − 2) = e ⇒ r ≅ 10 a = 5.5 fm 142 nearest value of A is Ba 10
r/a
15. A vessel, fitted with a weightless, frictionless 2 piston of 0.025 m …
Sol: Ideal gases cannot be liquefied as there is no attraction between the molecules. Real gases show ideal behavior at low pressure and high temperature. None of the gas is ideal.
19. The correct order for the rule of SN…….
Ans : 75% Ans : B> C > A> D
Sol: Work done = −P∆V 3 3 = −1 (atm) × 0.025 × 10 (dm ) Fe + 2HCl → FeCl2 + H2 Number of moles of Fe = Number of moles of H2 Work done = −nRT 1 1 = −n × 0.0821 (L atm mol− K− )× 300 K 25 ∴n= =1 0.0821 x 300
20. The aromatic species among the following are… Ans : D & E Sol: (D) is tropyllium cation which is aromatic according to Huckel’s rule. (E) is aromatic because is forms cyclobutenyl dication with 2 delocalised π electrons.
mass of Fe = 55.85 55.85 x 100 % purity = ≅ 75% 75 16. A solution at 298 K is separated from the pure solvent by a semi-permeable membrane. Difference in the height of the solution ……..
21. The reactions that give products with dipole moment are……… Ans : A, C & D
Ans : 250.20 K Sol: Osmotic pressure, π = hdg 3 2 = 0.9 × 1 × 10 × 9.8 N m− =
0.9 x 9.8 x 10 3 1.013 x 10 5
0.9 x 9.8 x 10 3 1.013 x 10 5 x 0.0821 x 298
∆Tf = Kf × m 3 = 30 × 3.5 × 10− = 0.1 ∴ F. P = 250.2 K
Electron current
17. Ans :
Sol: The product in (A) is cis but-2-ene. Reaction (C) gives pent-2-ene. All these products have dipole moments.
atm
22. The total number of stereoisomers for the following compound is…..
π = MST
M=
Sol: The relative stability of carbocation formed by hydrolysis is B> C > A> D
= 3.5 × 10−
3
Ans : 4 Sol: The compound contains two double bonds and the group linked to the two double bonds are different. Hence four stereoisomers are possible. They are E-E, Z-Z, E-Z and Z-E.
23. The correct order of ligand field strength is Ans : Cl− < H2O <NH3< CO
Photon energy
Sol: CO is a strong field ligand. The correct order of field strength is Cl− <H2O < NH3 < CO
24. The complex exhibiting a spin-only magnetic moment…….
28. Some gases in column X may be associated with options in Y.
Ans : Na2[Cr[NCS]4 [NH3]2]
Ans : SO2 → M, O, & R
Sol: In Na2[Cr[NCS]4 [NH3]2] the central atom is in +2 oxidation state and the hybridization is 2 3 d sp . There are two unpaired electrons.
Sol: SO2 is present in Troposphere. It is a component of classical smog. It is also responsible for acid rain. th
29. Let x be the 7 term from the beginning and y be th the 7 term …. 2
d sp Cr
3
Ans : 9
2+
n
Sol:
25. The most stable species among the following is Cl
Ans : Cl
n
Al
Al
Cl
Cl
n
x 1 = ⇒ y 12
Sol: Al2Cl6 is a stable species existing in solid state.
n
26. According to the VSEPR model, ……… ⇒ 3 X
r
n −r
1 Tr+1 = Cr . 3 3
Cl
Cl
1 1 x = T7 ; y = Tn−6; 3 3 + 1 4 3 4 1 4 3
1 C6 . 3 3
n− 6
1 . 1 4 3
1 1 Cn −6 . 3 3 . 1 4 3
n − 4 +n − 4 3 .4 3
=
6
n− 6
=
1 12
1 = 3 −1.4 −1 12
⇒n=9
Ans :
30. The number of functions f: {1, 2, …n} → {1, 2, …m}, where m, n, are positive….
X Sol: lone pair arrange at 120° to each other to form linear molecule.
27. Cyclohexanol is converted to Nylon-6 by……
Ans : m Sol:
Ans : Cu(250°C)/NH2OH/H /250°C O OH +
{1, 2, …n} → {1, 2, …m} f(1) = 1 ∴ for the remaining (n−1) elements there n 1 are m choices each ⇒ m −
31. The number of real roots of the equation in the interval (1/(x − 1)) + (1/(x − 2)) + …+1/(x − 5) = 1…
Cu/250°C
Sol:
n−1
NOH Ans : 4 H+
NH2OH
Sol:
O
250° C
NH Caprolactum
Nylon 6
In the question ‘in the interval’ is repeated twice. Please read after deleting the phrase the first time it appears 1 1 1 + + ... + =1 (x − 1) (x − 2) (x − 5) It is obvious that f(x) → −∞ when x → 1− , 2−, 3−, 4−, 5−. Also that f(x) → +∞ when + + + + + x → 1 , 2 , 3 , 4 , 5 and in (1, 2), (2, 3), (3, 4), (4, 5), f(x) is continuous.
2yy’ = 4 slope = (y') 1
4
( 4,1) = 2 = 2
1
2
3
4
Equation : tangent 1 ⇒ y − 1 = 2 x − 4
5
2(4 x − 1) 2 2y − 2 = 4x − 1 4x − 2y + 1 = 0 4k + 1 =2 5 16 + 4 4k + 1 =2 5 2 2 5 =
The approximate graph y = f(x) is given above. Therefore number of roots = 4 Aliter The given function can be modified to (x − 2) (x − 3) (x − 4)(x − 5) + (x − 1) (x − 3) (x − 4)(x − 5) + (x − 1) (x − 2) (x − 4)(x − 5) + (x − 1) (x − 2) (x − 3)(x − 5) + (x − 1) (x − 2) (x − 3) (x − 4) − (x − 1) (x − 2) (x − 3) (x − 4)(x − 5) = f(x)
4k + 1 = 4 × 5 4k = 20 − 1 = 19 19 k= 4
34. The point (−3, 2) undergoes the following … When x = 1, f(x) is +ve x = 2, f(x) is −ve x = 3, f(x) is +ve x = 4, f(x) is −ve x = 5, f(x) is +ve
1 3 Ans : , 2 2 Sol:
Since there are 5 sign changes, there are 4 roots. 32. Let y be the solution of the problem x y’ − y = 1 + 5e− , y(0) = y0 … 7 2
Ans : −
Sol:
The general solution is 5 x y = Ce − 1 − e − x 2 7 y(0) = y0 ⇒ C − = y0 2 7 C = y0 + 2 But given lim | y( x ) | is finite x →∞
⇒ C → 0 ⇒ y0 = −
7 2
33. The distance from a point (k, 0) on the positive xaxis to the tangent …. Ans :
19 4
Sol:
y = 4x
2
1
y =
4 2y
(−3, 2) Reflection about y = −x ⇒ (−2, 3) Reflection about x = 0 ⇒ (2, 3) Translation by 2 units along y ⇒ (2, 1) π Rotation by anticlock wise about origin 4 1 ⇒ (2 + i) × (1 + i) 2 1 = (1 + 3i) 2 1 3 ∴ The point is , 2 2
35. The equation of the plane passing through the intersection of the planes x + 2y + z − 1 = 0.. Ans : −4 Sol:
Equation : plane through intersection is (1 + 2λ)x + (2 + λ)y + (1 + 3λ)z − 1 − 2λ=0 1 + 2λ + 2 + λ + 1 + 3λ = 0 4 + 6λ = 0 −4 −2 λ= = 6 3 2 x + 2y + z − 1 − (2 x + y + 3z − 2) = 0 3 3x + 6y + 3z − 3 − 4x − 2y − 6z + 4 = 0 −x + 4y − 3z + 1 = 0 x − 4y + 3z − 1 = 0 ∴ k = −4
36. The principal value of 5π π 5π π 1 sin− sin cos + cos sin is … 9 9 9 9
Ans :
π 3
Sol:
5π π 6π 1 sin− sin + = sin−1 sin 9 9 9
1 2 0 39. Let P = − 1 1 2 and Q be such that … 1 2 1 Ans : real numbers
2π = sin−1 sin , 3 2π π π 1 i.e., sin− sin π − = sin−1 sin = 3 3 3
Sol:
Qx = λX ⇒ (Q − λI) X = 0 ⇒ |Q − λI| = 0 (Θ X is non zero) |P| = 1(1 − 4) − 2(−1 −2) = −3 + 6 = 3 − 3 − 2 4 1 Q = − 3 1 − 2 3 3 − 3 0 Q − λI =
37. Four numbers are chosen at random without replacement from the first 15 … 8
Ans : 1−
Sol:
C4
15
C4
Total number of ways of choosing 4 out of 15 15 = C4 For product to be even atleast are of 4 chosen numbers should be even. P(product even) = 1 − P(No even numbers) = 1−
8
C4
15
C4
38. Let M be a non-singular matrix of order 5 × 5…. Ans : M Sol:
−4
|adj M| = |M| 1 M−1 = M
|Q − λI| = −3(4 − 4 (1 − λ) + (3 − λ) [−(−3−λ) (1 − λ) −6] 2 = (1 − λ) (9 − λ + 6λ) = 0 ⇒ All real roots ⇒ λ ∈ R
x a a a a a x a a a
40. The value of determinant a a x a a is a a a x a a a a a x Ans : (x + 4a) (x − a) 1 1 1 1 1 a x a a a Sol: (x + 4a ) a a x a a a a a x a a a a x a 4
1 0 a x−a
4
(x + 4a ) a 4
|adj M− | = M−1 = 1
4 −3 − λ − 2 1 1− λ − 2 −3 3 0 3 − λ −3
1 M
4
=M
−4
a a
0 0
0 0
x−a 0 0 0 x−a 0 0 0 x−a
0 0 0
(x + 4a) (x − a)
0 0
4