# Common Physical Biological Sciences Practice Test 4

Question – 1

1. Consider the given figure in static equilibrium. What is the force, F required to hold the block of mass 100 Kg at rest? • A. 100 N

• B. 980 N

• C. 98 N

• D. 10.2 N

• E. Body will be at rest irrespective of any force applied

To find the force required to hold the block at rest, we need to equate the force, F acting downward with the tension acting upwards. Weight of the block can be found as
F = m x g ————(1)
where
F is the weight acting downwards;
m is the mass of the block;
g is the acceleration due to gravity;
Since, m = 100Kg and we will take g = 9.8 m/s2 putting the values in the equation (1) we get
F = 100 x 9.8
F = 980 N
Hence, some force has to be applied externally to keep the body at rest so E cannot be correct.
Tension, T equal to the weight of the block develops in the right string. Since, strings on both the sides are connected with each other they will experience same tension. Hence, tension in the left string equals to 980 N. As said before to keep the block in static equilibrium force required should equal the tension. Hence,
F = T
Or,
F = 980 N
Hence, b is the correct answer option.

Question – 2

2. Which of the following is a nucleophile?

• A. H+

• B. H3O+

• C. CO2

• D. AlCl3

• E. NH3

Generally, electron pair donors are called as nucleophiles. In the case of ammonia, it has a lone of pair of electrons. So, it can donate the electrons to the other atoms. But, the rest of the atoms can’t donate a pair of electrons instead they can accept a pair of electrons. So, except ammonia all the other molecules above are known as electrophiles.

Question – 3

3. Consider the given figure in static equilibrium. What is the tension T1 in the string? • A. 134.5 N

• B. 150 N

• C. 212 N

• D. 70.7 N

• E. 86 N

To find tension in the string we need to resolve the forces along x and y axis. Since the block is in equilibrium, equating the forces along, x and y axis we get,
Fx = 0
Fy = 0
Hence, T1cos450 = T2cos300
Or, T2 =
$\sqrt{\frac{2}{3}}$T1—————–(1)
Equating forces along y-axis we get
T1sin450 + T2sin300 = 150 ———–(2)
Putting the value of T2obtained in equation (1) in (2) and solving for T1we get
T1sin450 +

$\sqrt{\frac{2}{3}}$T1sin300 = 150
Or,

+

$\sqrt{\frac{2}{3}}$T1 x

$\frac{1}{2}$ = 150
Or, T1 = 134.5 N
Hence, a is the correct answer option.

Question – 4

4. Which of the following develops from ectoderm?

• A. Liver

• B. Pancreas

• C. Kidney

• D. Stomach

• E. Brain

Liver, pancreas and stomach develop from endoderm. Kidney develops from intermediate mesoderm. Brain develops from the neural tube of the ectoderm. Ectoderm generates the outer layer of the embryo.The ectoderm develops into the surface ectoderm, neural crest and neural tube.

Question – 5

5. Arrange the following in the increasing order of acidity • A. B > C > A

• B. C < B < A

• C. A < B < C

• D. C < A < B

• E. A < C < B

The order of acidity can be identified by inductive effect. The 2-chlorobutanoic acid (molecule C) has greater inductive effect than 3-chlorobutanoic acid (B) and 4-chlorobutanoic acid (A). So, 2-chlorobutanoic acid (molecule C) is the strongest acid than 3-chlorobutanoic acid and 4-chlorobutanoic acid. The increasing order of acidity is A < B < C.

Question – 6

6. An infection of placenta is known as

• A. Chorioamnionitis

• B. Hepatitis

• C. Laminitis

• D. Mastitis

• E. Uveitis

An infection of the placenta is known as chorioamnionitis. Bacteria enter through the birth canal or reaching the placenta through blood stream. Symptoms are fever, abdominal tenderness and rapid heart rate.

Question – 7

7. Consider the given figure in static equilibrium. What is the tension T2 in the string? • A. 100 N

• B. 173 N

• C. 200 N

• D. 115 N

• E. 58 N

To find tension in the string we need to resolve the forces along x and y axis. Since the block is in static equilibrium, equating the forces along x and y axis as
FX = 0
FY = 0 Therefore,T2cos600 = T1cos300
Or, T1 =

$\frac{1}{\sqrt{2}}$T2——–(1)
Equating the forces along y-axis we get
T1sin300 + T2sin600 = 200—————(2)
Putting the value of T1 obtained in equation (1) in (2) and solving for T2 we get

$\frac{{T}_{2}}{\sqrt{3}}??\frac{1}{2}+\frac{{T}_{2}??\sqrt{3}}{2}$ = 200

$\frac{2{T}_{2}}{\sqrt{3}}$ = 200
Or, T2 = 173 N
Hence, b is the correct answer option.

Question – 8

8. Which of the following functional groups shows characteristic peak around 2100-2260 cm-1 in IR spectroscopy?

• A. Ketone

• B. Alcohol

• C. Alkyne

• D. Alkene

• E. aldehyde

Alkynes show the characteristic peak around 2200cm-1 in IR spectroscopy. Carbonyl compound’s characteristic peak appears at 1680-1750 cm-1. Alcohol shows the characteristic peak at 3200-3500 cm-1.Alkenes show the characteristic peak at 1680-1600 cm-1.

Question – 9

9. A 6-N and 8-N force acts on a point concurrently. Which of the following force cannot produce a state of equilibrium?

• A. 14N

• B. 2N

• C. 10N

• D. 12N

• E. 16N

There can be four possible orientations of these two forces.
1) If both of them are acting along same direction. In this case both the forces add together to give maximum effect which is 14N. Hence, force equal to 14N but acting in opposite direction will produce a state of equilibrium. Therefore, a is incorrect.
2) If both the forces are acting in opposite direction. In this case both the forces cancel out each other to provide net force of 2N. Hence, force equal to 2N but acting in opposite direction will produce a state of equilibrium. Therefore, b is incorrect.
3) If both the forces are orthogonal to each other. In this case net force can be calculated as
R =
$\sqrt{{8}^{2}+{6}^{2}}$R = 10 N Hence, force equal to 10N but acting in opposite direction as shown above will produce a state of equilibrium. So, c is incorrect.
4) If both the forces are acting at some angle. In this case net force can be calculated by applying vector’s addition formula.
R =

$\sqrt{{6}^{2}+{8}^{2}+2??6??cos??}$Where, R is the resultant of the two forces as shown above;
?? is the angle between the two forces;
Since, cos?????1
Hence R is less than 14 N. Therefore, any force greater than or equal to 2N and any force less than or equal to 14N can balance these two forces. Hence, option d is incorrect. Any force less than 2N and more than 14N cannot produce a balancing effect. Hence, e is the only correct answer option.

Question – 10

10. In which of the following estrous phases is a female animal sexually receptive?

• A. Proestrus

• B. Estrus

• C. Metestrus

• D. Diestrus

• E. Anestrus

A female animal shows a sexually receptive behavior which may be visible by certain physiological changes due to gonadotropic hormones. In some species ovulation may occur. This phase of estrous cycle is called estrus.

Question – 11

11. Consider the following statements.1) Body in equilibrium, has uniform velocity.2) Body in equilibrium, has non-zero acceleration.

• A. Both the statements are true

• B. Both the statements are false

• C. Statement 1 is true but 2 is false

• D. Statement 2 is true but 1 is false

• E. Statement 2 is correct explanation for the statement 1

For a body in equilibrium, there may be two cases associated with it.
1)In equilibrium all the forces cancel out each other and hence body stays at rest forever unless and until some unbalanced force acts on it.
2)A body may be in equilibrium if it is moving with uniform velocity. Since, acceleration is the time rate of change of velocity, constant velocity results in zero acceleration.
Therefore, statement 1 is correct but 2 is incorrect hence c is correct answer option.

Question – 12

Molecule Bond order
Nitrogen 3
Oxygen 2
Lithium 1
Nitric oxide 2.5

• A. Lithium

• B. Oxygen

• C. Nitric oxide

• D. Nitrogen

• E. Both c & d

Bond length is found to be inversely proportional to the bond order. Greater the bond order, shorter the bond length and vice versa. From the given data, the bond order for lithium is one, so its bond length will be greater than for the other molecules.

Question – 13

13. Which of the following can prevent the differentiation of stem cells?

• A. Leukemia inhibiting factor

• B. Bone morphogenetic proteins

• C. IL-3

• D. TNF

• E. Both a and b

Both leukemia inhibiting factor and bone morphogenetic factors can be used to prevent the differentiation of stem cells. These factors are important and useful for deriving embryonic stem cells. IL-3 and TNF (tumour necrotic factor) are cytokines.

Question – 14

14. Consider a block is kept on a horizontal floor as shown below. A force of 25N making an angle of 60o from the horizontal pulls it with uniform velocity. What is the value of frictional force acting against the force? • A. 25N

• B. 20N

• C. 12.5N

• D. 50N

• E. 29N

Since the block is undergoing uniform velocity no acceleration in it takes place. Resolve the forces acting on the body along x and y axis. br/>Where
Fr is the frictional force;
25 cos600 is the horizontal component;
25 sin600 is the vertical component;
Since, body moves with uniform velocity by equating the horizontal forces we get
25cos600 = Fr
Or,
$\frac{25}{2}$ = Fr
Or, Fr = 12.5 N
Therefore, c is the correct answer option.

Question – 15

15. Which of the following syndrome is caused due to the presence of an extra X chromosome in humans?

• A. Klinefelter’s syndrome

• B. Cri du chat syndrome

• C. Turners syndrome

• D. Edwards Syndrome

• E. Down’s Syndrome