Jee Knockout Mock Test

8/10/2019 Jee Knockout Mock Test

1/15


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2

5 2 22 32000 10 cm 3000 10 cm 1.9 10 cm .2

V

Hence the correct option is (1).

Measurement of time

2. A government once attempted to base measures of time on multiples of ten (decimal system):

One week consisted of 10 days, one day consisted of 10 hours, one hour consisted of 100minutes, and one minute consisted of 100 seconds. What is the ratio of this decimal week to thestandard week?

(1) 7:10 (2) 10: 7(3) 5:2 (4) 2:5

Solution:

Presuming that a decimal day is equivalent to a regular day, then the ratio of weeks is simply 10/7 or (to 3significant figures) 1.43.


Measurement of time

3. Because Earth's rotation is gradually slowing, the length of each day increases: The day at the endof 1.0 century is 1.0 ms longer than the day at the start of the century. In 20 centuries, what is the

total of the daily increases in time?(1) 7305 s (2) 3057 s

(3) 3456 s (4) 6543 s

Solution:

The last day of the 20 centuries is longer than the first day by

20 century 0.001 s century 0.02 s. The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day. Since theincrease occurs uniformly, the cumulative effect Tis

average increase in length of a day number of days

0.01s 365.25 day2000 y

day y

7305 s

T


Measurement of mass

4. A vertical container with base area measuring 14.0 cm by 17.0 cm is being filled with identical

pieces of candy, each with a volume of 50.0 mm3and a mass of 0.0200 g. Assume that thevolume of the empty spaces between the candies is negligible. If the height of the candies in the

container increases at the rate of 0.250 cm/s, at what rate (kilograms per minute) does the mass ofthe candies in the container increase?(1) 4.31 kg / min (2) 3.41 kg/min


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(3) 2.34 kg/min (4) 1.43 kg/min

Solution:

The mass density of the candy is

3

-4 3

-4 3

=

0.0200g=

50.0 mm

= 4.0010 g / mm

= 4.0010 kg / cm

m

V

If we neglect the volume of the empty spaces between the candies, then the total mass of the candies in

the container when filled to height his ,M Ah where 2(14.0 cm)(17.0 cm) 238 cmA is the base

area of the container that remains unchanged. Thus, the rate of mass change is given by

4 3 2

( )

(4.00 10 kg/cm )(238 cm )(0.250 cm/s)

0.0238kg/s

1.43kg/min

dM d Ahdt dt

dhA

dt


Conversion between units

5. In old rural England 1 hide (between 100 and 120 acres, where 1 acre is equal to 4047 m2) was

the area of land needed to sustain one family with a single plough for one year. Also, 1 wapentake

was the area of land needed by 100 such families. In contrast, in quantum physics the cross-sectional area of a nucleus is measured in units of barns, where 1 barn is 1 1028 m2. What isthe ratio of 25 wapentakes to 11 barns?

(1) 361 10 (2) 362 10

(3) 301 10 (4) 302 10

Solution:

Abbreviating wapentake as wp and assuming a hide to be 110 acres, we set up the ratio 25 wp/11 barnalong with appropriate conversion factors:

2

28 2

36

100 hide 110acre 4047 m

1 wp 1acre1hide

1 10 m

1barn

25 wp

1 10 .

11 barn



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Dimensions

8. Let 0[ ] denote the dimensional formula of the permittivity of vacuum. IfM= mass,L= length,

T = time andA= electric current, then:

(1) 1 3 4 20[ ] [ ]M L T A (2) 1 2 1 20[ ] [ ]M L T A

(3) 1 2 10[ ] [ ]M L T A (4) 1 3 20[ ] [ ]M L T A

Solution:

According to Coulombs law the magnitude of force between two charges of magnitude qand separated

by distance r is given by relation2

20

2

0 2

1

4

1

4

qF

r

q

Fr

Substituting the dimensions of q= AT , 2

=F MLT

, and r L

we get the dimensional formula of

the permittivity of vacuum as2 2

1 3 2 40 2 2

[ ][ ]

[ ][ ]

A TM L A T

MLT L

Hence, the correct option is (1).

Significant digits

9. The respective number of significant figures for the numbers 23.023, 0.0003, and3

2.1 10 are(1) 4, 4, 2 (2) 5, 1, 2(3) 5, 1, 5 (4) 5, 5, 2

Solution:

Number of significant figures in 23.023 is 5.

Number of significant figures in 0.0003 =43 10 is 1.

Number of significant figures in 32.1 10 is 2.Hence the correct option is (2).

Error analysis

10. The current voltage relation of diode is given byI = (e1000V/T

1) mA, where the applied V is involts and the temperature T is in degree kelvin. If a student makes an error measuring 0.01 V

while measuring the current of 5 mA at 300 K, what will be the error in the value of current inmA?

(1) 0.2 mA (2) 0.02 mA(3) 0.5 mA (4) 0.05 mA

Solution:The given current voltage relation is

I = (e1000V/T1) mA. (1)


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Now atI= 5 mA, Eq. (1) becomes

5= (e1000V/T1)or e1000V/T= 6 mA

Differentiating Eq. (1) w.r.t V, we get

1000 /

1000 /

1000

1000

10006mA 0.01

300

0.2mA

V T

V T

dI

edV T

dI e dV T


Part BChemistry

Basic Stoichiometry-I (Some Basic Concepts in Chemistry)

Mole concept

1. Set the following in increasing order of mass: (a) 4.55 1022

atoms of Pb, (b) 8.50 g of C, (c)

7.14 1022

atoms of Zn, (d) 0.280 mol of Ca(1) (a) < (c) < (d) < (b) (2) (b) < (d) < (a) < (b)(3) (d) < (b) < (c) < (a) (4) (a) < (b) < (c) < (d)

Solution:

Converting all the quantities into mole, we get

(a) Number of moles of Pb =

22

234.55 10 0.0755 mol6.023 10

(b) Number of moles of C =8.50

0.708 mol12.0

(c) Number of moles of Zn =22

23 0.118 mol

6.023

7.14 10

10

(d) Number of moles of Ca = 0.280 mol.

Therefore, the increasing order is (a) < (c) < (d) < (b).Hence, the correct option is (1).

Mole concept

2. From 392 mg of H2SO4, 1.204 1021 molecules are removed. How many moles of H2SO4 are

left?

(1) 2.0 103 (2) 1.2 10

3

(3) 4.0 103 (4) 1.5 10

3

Solution:

Given weight of H2SO4= 392 mg


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6.023 1023molecules of H2SO4= 98 g

Therefore, 1.204 1021molecules = 21

23 1.204 1

98g 0.0

6.02196 g 196 mg

3 10

Now, weight of H2SO4left = Given weight Weight removed = 392 196 = 196 mg3

3WeightSo,the moles left

Molecul

196 10

a

2.0 10

9r weight 8


Empirical and molecular formula determination

3. If 30 mL of a gaseous hydrocarbon requires 90 mL of O2for complete combustion and 60 mL of

CO2is formed in the process, the molecular formula of hydrocarbon will be

(1) C2H2 (2) CH4(3) C2H4 (4) C2H6

Solution:

Let the hydrocarbon be CxHy, then it undergoes combustion in the following manner:

2 2 24 2C H O CO H O30 mL 90 mL 60 mL

1mL 3mL 2 mL

or 1 mol 3 mol 2 mol

y yxx y x

Equal volumes of all gases contain equal number of molecules at the same conditions of temperature and

pressure. Therefore, we have

2 and 3 3 2 1 44 4

y yx x y

Thus, the molecular formula = C2H4.


Concentration termsMolarity

4. The concentration of bivalent lead ions in a sample of polluted water that also contains nitrate

ions is determined by adding solid sodium sulphate (M = 142 g mol1) to exactly 500 mL of

water. Calculate the molarity of lead ions if 0.355 g of sodium sulphate was needed for completeprecipitation of lead ions as sulphate.

(1) 1.25 103

M (2)25 103

M

(3) 5 103

M (4)None of these

Solution:

The reaction involved is

3

2 4 3 2 4 30.355 g

2.5 10 mol

Na SO Pb(NO ) PbSO 2NaNO

Hence, the number of moles of Na2SO43

2 4

0.355g2.5 10 mol of Na SO

142

Thus, molarity of lead ions is


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332.5 10 0.005M 5 10 M

0.5


Interconversion of concentration terms

5. The density of 1 M solution of NaCl is 1.0585 g mL1. Which of the statements is true?

(a) The molality of the solution is 1.0585 M(b) The molarity of the solution is greater than molality

(c) With increase in temperature, the molality will increase(d) The molarity of the solution is equal to molality.

(1) c and d (2) b and c(3) a only (4) d only

Solution:

(a) is incorrect. The molality is given by

B1 1 1.0585 58.5 1 molal1000 1 1000

Mdm

m M m

where d= density of solution, m= molality,M= molarity andMB= molar mass of solute

(b) is incorrect(c) is incorrect as molality is independent of temperature.

(d) is correct for this case.


Percentage composition in compounds

6. The correct order of mass percentage of elements in ferric sulphate is

(1) Fe > S > O (2) S > O > Fe(3) O > Fe > S (4) Fe > O > S

Solution:

The formula of ferric sulphate is Fe2(SO4)3. Hence,

2 Fe, 3 S, 12 O

2 56, 3 32, 12 16112, 96, 192O > Fe > S


Stoichiometry of reactionsCalculation based on chemical equations

7. Consider the following reaction: 2 3CO(g) 2H (g) CH OH(g) . Which of the following statements

is true?(a) 4 mol of H2are required to react completely with 2 mol of CO.

(b) 1.2046 1024H2molecules would be needed to consume 2 mol of CO.

(c) 2.41 1024H atoms would be needed?

(d) 8 g of H2are required to consume 2 mol of CO?


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(1) a and b (2) a, b, c

(3) a and d (4) b, c, d

Solution:

(a) 2 2 = 4 mol of H2are needed to consume 2 mol of CO.

(b) 4 6.023 1023= 2.41 1024molecules would be needed to consume 2 mol of CO

(c) 2 2.41 1024= 4.82 10

24atoms will be needed to consume 2 mol of CO.

(d) The molar mass of H2is 2.0 g mol1

. Four moles of H2will have a mass of 4 2 = 8 g.Hence, the correct option is (3).

Concept of limiting reagent

8. A sample of calcium oxide whose mass was 12 g was treated with 46.0 g of water to prepare

calcium hydroxide. Which compound was the limiting reactant?(1) Ca(OH)2 (2) CaO

(3) O2 (4) H2O

Solution:

The reaction equation is2 2CaO H O Ca(OH) . From the equation, we have

2

12 4612 g of CaO 0.23mol 46.0 g of H O 2.55 mol

56.1 18

It follows that one mole of CaO reacts with one mole of H 2O. Therefore, H2O is present in excess and

some of it will remain unreacted when the reaction is over. Hence, CaO is the limiting reagent.


Percentage yield of reaction

9. Cyclohexanol is dehydrated to cyclohexene on heating with conc. H 2SO4. The cyclohexene

obtained from 100 g cyclohexanol will be (if yield of reaction is 75%)

(1) 61.5 g (2) 75.0 g(3) 20.0 g (4) 41.0 g

Solution:

One molecule of cyclohexanol should produce one molecule of cyclohexene. Now, molecular weight of

cyclohexanol = 100.2 g mol1, molecular weight of cyclohexene = 82.1 g mol

1. Therefore,100.2 g of cyclohexanol gives = 82.1 g of cyclohexene

100 g of cyclohexanol82.1 1

gives 00

1 .2

00

Since the yield of reaction is 75 %, the amount of cyclohexene formed is

82.1 100 7561.5g

100.2 100



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Calculation based on chemical equations

10. An experiment requires 100 cm3of 20.0% H2SO4, density 1.14 g cm

3. How much concentrated

acid of density 1.84 g cm3and containing 98% H2SO4by weight must be diluted with water to

prepare 100 cm3acid of the required solution?

(1) 8.1 cm3 (2) 21.3 cm

3

(3) 18.1 cm3 (4) 12.7 cm3

Solution:

Required for experiment = 100 cm3of 20% H2SO4; d= 1.14 g cm3

%Acid Density 10 20 1.14 10Strength 2.33M

Molecular Mass 98

Given that % H2SO4= 98%, d= 1.84 g cm3

98 1.84 10Strength 1.84M

98

Volume of the H2SO4of conc. 18.4 M required is given by substitutingM1= 2.33 M,M2= 18.4 M

1 12

2

2.33 100 12.66mL 12.7mL18.4

M VVM


Part CMathematics

Quadratic Equations and Expressions

Root or solution of quadratic equation: Real roots

1. If the roots of the equationsx2bx+ c= 0 andx2cx+ b= 0 differ by the same quantity, then b

+ cis equal to(1) 4 (2) 1

(3) 0 (4)4

Solution:

We know that if and are roots of the equation2 0ax bx c

then2 4b ac

a

Equating the value of from both the given equations, we get

2 2 2 24 4 4 4b c c b b c c b 2 2

4( ) ( )( 4) 0

4 ( )

b c b c b c b c

b c b c



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Formation of equation: With the help of roots of other equation

2. The value of a, for which one root of the quadratic equation 2 2( 5 3) (3 1) 2 0a a x a x is

twice as large as the other, is

(1)2/3 (2) 1/3

(3)1/3 (4) 2/3

Solution:

Let us consider that and 2be the roots of the given quadratic equation. Then2 2

( 5 3) (3 1) 2 0a a a (1)

and 2 2( 5 3)(4 ) (3 1)(2 ) 2 0a a a (2)

Multiplying Eq. (1) by 4 and subtracting it from Eq. (2), we get

(3 1)(2 ) 6 0a

It is obvious that a1/3. Therefore,

3

3 1a

Putting this value in Eq. (1) we get

2 2 2( 5 3)(9) (3 1) (3) 2(3 1) 0a a a a

2 29 45 27 (9 6 1) 0 39 26 0a a a a a

2

3a

Forx= 2/3, the equation becomesx2+ 9x+18 = 0 whose roots are3 and6.


Formation of equation: When roots are given

3.

If aand bare rational and andbe the roots ofx2+ 2ax+ b= 0, then the equation with rational

coefficients, one of whose roots is2 2

,is

(1) 2 4 2 0x ax b (2) 2 4 2 0x ax b

(3) 2 4 2 0x ax b (4) 2 4 2 0x ax b

Solution:

Given that andare the roots ofx2+ 2ax + b = 0, we have

2 anda b

Let us consider that

2 2y

which implies that


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2 2 2 2 2

2

( 2 ) ( ) 2 4 2

4 2 0

y a a b

y ay b

Therefore, the required equation is

2+ 4 + 2 = 0x ax b


Root or solution of quadratic equation: Imaginary roots

4. If the roots of the equation bx2 + cx + a = 0 be imaginary, then for all real values of x, theexpression 3b2x2+ 6bcx + 2c2is

(1) greater than 4ab (2) less than 4ab(3) greater than4ab (4) less than4ab

Solution:

The given equation is2 0bx cx a

Considering the roots are to be imaginary, we have2

2

2

4 0

4

4

c ab

c ab

c ab

Therefore,

2 2 23 6 2b x bcx c

as 3b2> 0. Given expression has minimum value, which is obtained as follows:

2 2 2 2 2 22

2 2

4(3 )(2 ) 36 124

4(3 ) 12

b c b c b cc ab

b b


Root or solution of quadratic equation: Real roots

5. If the difference between the roots of the equationx2+ ax+ 1 = 0 is less than 5, then the set of

possible values of ais

(1) (3, 3) (2) (3, )(3) (3, ) (4) (, 3)

Solution:2

1 0x ax


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2

2

2

2

2

1

( ) 4

4

4 54 5

9 0

( 3,3).

a

a

aa

a

a


Relation between roots and coefficients in quadratic equation: Discriminant of quadratic equation

6. If 2 2 2sin cos sin , then the roots of the equation 2 2 cot 1 0x x are always

(1) equal (2) imaginary

(3) real and distinct (4) greater than 1Solution:

The discriminant of the given equation is2 2

2

2 2 2

2

2

2

4 cot 4 4(cot 1)

4(cosec 2)

4 48 8

sin sin cos

28 1

sin 2

2 sin 28 0

sin 2


Theory of equation: Use of Rolles theorem

7. The number of values of kfor which the equationx

22x+ k= 0 has two distinct roots lying in

the interval (0, 1) is(1) 0 (2) 1

(3) 2 (4) infinitely many

Solution:

Let the two distinct roots lying between 0 and 1 be andsuch that


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Sincef(x) =x22x+ k is a differentiable andf() =f() = 0, by the Rolles theorem, there exist

( , ) (0,1) such that

( ) 0

2 2 0

1

f

Therefore, two distinct roots lying in the interval (0, 1) is not possible.


Theory of equation: Remainder theorem

8. Letxbe an integer andx

2+x+ 1 is divisible by 3. Whenxis divided by 3, it leaves the remainder

(1) 0 (2) 1(3) 2 (4) any of (1), (2) and (3)

Solution:

Let us consider thatx= 3m+ r, where 0 r 2. Therefore,2 2 21 (9 6 3 ) ( 1)x x m mr m r r

Note thatx2+x+ 1 is divisible by 3 if and only if r= 1.


Analysis of quadratic expression: Sign of quadratic expression

9.

Letf (x) be a quadratic expression such that f (x) < 0 x.

If ( ) ( ) ( ) ( )g x f x f x f x thenfor ,x

(1) ( ) 0g x (2) ( ) 0g x

(3) ( ) 0g x (4) ( ) 0g x

Solution:

Letf(x) = ax2+ bx+ c2

( ) 0 0, 4 0f x x a b ac

Now,2

( ) (2 ) 2g x ax bx c ax b a 2

(2 ) (2 )ax a b x a b c

Herethe discriminant is (2a+ b)24a(2a+ b+ c)

Using2

0 and 4 0a b ac , we get the discriminant as,

2 2( 4 ) 4 0b ac a


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Since discriminant of g x is less than 0 and 0, 0a g x .


Analysis of quadratic expression: Maximum and minimum value of quadratic expression

10. If and are the roots of the equation x2 + ax + b = 0, then the maximum value of

2 21( )

4x ax b is

(1) 21

( 4 )4

a b (2) 21

( 4 )4

b a

(3)2

2

a (4) none of these

Solution:

2 2

22 2

2 22

1( ) ( ) ( )

4

( ) 4 ( )

4 2 4

1( )

2 2 2

f x x x

x

a ax

Thus, the maximum valuef(x) is2

2

a, which is achieved when .

2

ax


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