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Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the drawn ticket has a number which is a multiple of 3 or 7?
Favourable cases = 3, 6, 9, 12, 15, 18, 7, 14 = 8 cases
Required Probability = 8/20 = 2/5
Favourable cases = 3, 6, 9, 12, 15, 18, 7, 14 = 8 cases
Required Probability = 8/20 = 2/5
In a zoo, there are 42 animals in one sector, 34 animals in the second sector and 20 animals in the third sector. Out of these, 24 graze in sector one and also in sector two, 10 graze in both sector two and sector three and 12 graze in both sector one and sector three. These figures also include four animals who graze in all the three sectors. If all the animals are now transported to another zoo, find the total number of animals transported.
Let I, II, III be the sets of animals grazing in 1^{st}, 2^{nd} and 3^{rd} sectors of zoo respectively.
Total number of animals = 10 + 20 + 4 + 8 + 4 + 6 + 2 = 54
Let I, II, III be the sets of animals grazing in 1^{st}, 2^{nd} and 3^{rd} sectors of zoo respectively.
Total number of animals = 10 + 20 + 4 + 8 + 4 + 6 + 2 = 54
If x cos y + y cos x = π and initial value of x and y is equal to 0, then y’ (0) =
x cosy + y cosx = π
Differentiating both side with respect to x
cosy + x (siny) dy/dx + cosx dy/dx – y sinx = 0
dy/dx = (y sinx cosy)/(cosx – x siny)
Putting here x= 0 and y = 0
dy/dx = (1)/1 = 1
x cosy + y cosx = π
Differentiating both side with respect to x
cosy + x (siny) dy/dx + cosx dy/dx – y sinx = 0
dy/dx = (y sinx cosy)/(cosx – x siny)
Putting here x= 0 and y = 0
dy/dx = (1)/1 = 1
If f(x) = 3x – 5 and f(g(x)) = 2x, then find the value of g(x).
Given that f(x) = 3x – 5 and f (g(x)) = 2(x)
∴ 3. G(x) – 5 = 2x
∴ g(x) = (2x + 5)/3.
Given that f(x) = 3x – 5 and f (g(x)) = 2(x)
∴ 3. G(x) – 5 = 2x
∴ g(x) = (2x + 5)/3.
If x + y = K is normal to y^{2} = 12x, then K is
We know that x + y = K (slope = – 1) is normal to the given curve, it means it is perpendicular to the tangent at the point (x, y). Hence the slope of the tangent must be 1.
y = mx+c is a tangent to y^2 = 4ax, iff. c=a/m.
Putting a = 3, m =1, we get c = 3.
Tangent is y=x+3.
Solving the equations of parabola and tangent, we get pts (3,6) and(3,6)
This means that k= – 3 or k = 9
We know that x + y = K (slope = – 1) is normal to the given curve, it means it is perpendicular to the tangent at the point (x, y). Hence the slope of the tangent must be 1.
y = mx+c is a tangent to y^2 = 4ax, iff. c=a/m.
Putting a = 3, m =1, we get c = 3.
Tangent is y=x+3.
Solving the equations of parabola and tangent, we get pts (3,6) and(3,6)
This means that k= – 3 or k = 9
The smallest interval [a, b], such that dx ∈ [a, b] is given by
Choose the odd one out.
Correct Answer : Map study
Correct Answer : Map study
Match List I with List II and select the correct answer using the codes given below the lists:
List I  List II 
A. Camber  1. Subsurface drainage 
B. Joints  2. Surface drains 
C. Tack coat  3. Transverse slope 
D. Sand drains  4. Bituminous roads 
5. Concrete roads 
Correct Answer : A – 3, B – 5, C – 4, D – 1
Correct Answer : A – 3, B – 5, C – 4, D – 1
Which one of the following pairs in not correctly matched?
Correct Answer : Check valve – To check water flow in all directions
Correct Answer : Check valve – To check water flow in all directions
The atmosphere extends upto a height of 10,000 km. It is divided into the following four thermal layers:
1. Mesosphere 2. Stratosphere 3. Thermosphere 4. Troposphere
The correct sequence of these layers starting from the surface of the earth upwards is
Correct Answer : 4, 2, 1, 3
Correct Answer : 4, 2, 1, 3
Pick up the correct statement:
Correct Answer : Confined aquifers are also known as flowing wells.
Correct Answer : Confined aquifers are also known as flowing wells.
Pick up the correct statement from the following.
Correct Answer : The water supply pipes carry pure water free from solid particles.
Correct Answer : The water supply pipes carry pure water free from solid particles.
In Chezy’s formula, V = C √rs is used for calculating the velocity of flow in circular sewer running full. The value of hydraulic mean radius may be taken as
Correct Answer : D/4
Correct Answer : D/4
In New mark’s influence chart for stress distribution, there are ten concentric circles and ten radial lines. The influence factor of the chart is:
The consistency index of a soil is defined as the ratio of:
Correct Answer : liquid limit minus the natural water content to the plasticity index of the soil
Correct Answer : liquid limit minus the natural water content to the plasticity index of the soil
Which of the following conditions must be obeyed by the topmost flow line through an earthen dam section?
1. Having atmospheric pressure
2. Normal to the downstream slope
3. Normal to the upstream slope
4. Tangent to the slope at the exit point
Correct Answer : 2 and 4
Correct Answer : 2 and 4
Pick up the correct statement from the following.
Correct Answer : All the above
Correct Answer : All the above
The run off a drainage basin is
Correct Answer : Precipitation – ground water accretion – initial recharge
Correct Answer : Precipitation – ground water accretion – initial recharge
To estimate the magnitude of a flood with a return period of T years, Gumbel`s distribution method requires the following data pertaining to annual food series:
(i) mean value
(ii) standard deviation
(iii) length of record
(iv) coefficient of skew
Correct Answer : (i), (ii) and (iii)
Correct Answer : (i), (ii) and (iii)
The ordinate of the Instantaneous Unit Hydrograph (IUH) of a catchment at any time t, is
Correct Answer : the slope of the S curve with effective rainfall intensity of 1 cm/hr
Correct Answer : the slope of the S curve with effective rainfall intensity of 1 cm/hr
Which of the following will prove to be the most economical section of a circular channel for the maximum discharge?
Correct Answer : All the above
Correct Answer : All the above
In a standing wave flume, the depth of flow in the throat region
Correct Answer : should be equal to the critical depth
Correct Answer : should be equal to the critical depth
The relationship:
dp/dx = dr/dy is valid for
Correct Answer : uniform flow
Correct Answer : uniform flow
Match List I with List II and select the correct answer using the codes given below the lists:
List I  List II 
A. Rotational flow  1. A fluid motion in which stream lines are concentric circles 
B. Vortex flow  2. The fluid particles moving in concentric circles may not rotate about their mass centre 
C. Free vortex  3. The fluid particles moving in concentric circles may rotate 
D. Forced vortex  4. Flow near a curved solid boundary 
Correct Answer : A – 4, B – 1, C – 2, D – 3
Correct Answer : A – 4, B – 1, C – 2, D – 3
In steady laminar flow of a liquid through a circular pipe of internal diameter D, carrying a constant discharge, the hydraulic gradient is inversely proportional to:
Correct Answer : D^{2}
Correct Answer : D^{2}
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JEE Main Physics Gravitation Online Test. JEE Main Online Test for Physics Gravitation. JEE Main Full Online Quiz for Physics Gravitation. JEE Main Free Mock Test Paper 2019. JEE Main 2019 Free Online Practice Test, Take JEE Online Test for All Subjects. JEE Main Question and Answers for Physics Gravitation. In this test You may find JEE Main all subjects Questions with Answers. Check JEE Main Question and Answers in English. This mock Test is free for All Students. Mock Test Papers are very helpful for Exam Purpose, by using below Mock Test Paper you may Test your Study for Next Upcoming Exams. Now Scroll down below n Take JEE Main Physics Gravitation…
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Figure shows the variation of energy with the orbit radius r of a satellite in a circular motion. Mark the correct statement.
Kinetic energy KE = GMm/2r
Potential energy PE = – GMm/r
And the total energy E = – GMm/2r
Kinetic energy is always positive and KE ∝ 1/r
Potential energy is negative and PE ∝ 1/r
Similarly total energy is also negative and E ∝ 1/r
Also E < PE
∴ A is kinetic energy, B is potential energy and C is total energy of the satellite.
Kinetic energy KE = GMm/2r
Potential energy PE = – GMm/r
And the total energy E = – GMm/2r
Kinetic energy is always positive and KE ∝ 1/r
Potential energy is negative and PE ∝ 1/r
Similarly total energy is also negative and E ∝ 1/r
Also E < PE
∴ A is kinetic energy, B is potential energy and C is total energy of the satellite.
A planet revolving around sun in an elliptical orbit has a constant
In elliptical orbit un is at one of the foci hence the distance between the planet and sun changes as planet revolves hence linear speed, kinetic energy and potential energy of planet do not remain constant
In elliptical orbit un is at one of the foci hence the distance between the planet and sun changes as planet revolves hence linear speed, kinetic energy and potential energy of planet do not remain constant
Both earth and moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon
Moon is revolving around earth in almost circular orbit. Sun exerts gravitational pull on both, earth and moon. When observed from sun, the orbit of the moon will not be strictly elliptical because they total gravitational force (i.e., force due to earth on moon and force due to sun on moon) is not central.
Moon is revolving around earth in almost circular orbit. Sun exerts gravitational pull on both, earth and moon. When observed from sun, the orbit of the moon will not be strictly elliptical because they total gravitational force (i.e., force due to earth on moon and force due to sun on moon) is not central.
Three particles each having a mass of 100 g are placed on the vertices of an equilateral triangle of side 20 cm. The work done in increasing the side of this triangle to 40 cm is
W = U_{f} – U_{i}
m = 0.1 kg, r_{f} = 0.4 m and r_{i} = 0.2 m
Substituting the value, we get
W = 5.0 × 10^{12} J
W = U_{f} – U_{i}
m = 0.1 kg, r_{f} = 0.4 m and r_{i} = 0.2 m
Substituting the value, we get
W = 5.0 × 10^{12} J
Two satellites S_{1} & S_{2} of equal masses revolve in the same sense around a heavy planet in coplanar circular orbit of radii R & 4R
Two identical spherical masses are kept at some distance as shown. Potential energy when a mass m is taken from surface of one sphere to the other
Centre point is the unstable equilibrium position where potential energy is maximum.
Centre point is the unstable equilibrium position where potential energy is maximum.
If W_{1}, W_{2} and W_{3} represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively (as shown) in a gravitational field of point mass m then
Since the gravitational field is conservative field hence, the work done in taking a particle from one point to another in a gravitational field is path independent.
Since the gravitational field is conservative field hence, the work done in taking a particle from one point to another in a gravitational field is path independent.
Maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from earth’s surface is:
Applying energy conservation
⇒ R + h = 4h ⇒ h = R/3
Applying energy conservation
⇒ R + h = 4h ⇒ h = R/3
A thin spherical shell of mass M and radius R has a small hole. A particle of mass m is released at the mouth of the hole. Then
Net gravitational field inside a shell is zero. Hence, net force on the particle will be zero i.e., the particle stays at rest in its original position.
Net gravitational field inside a shell is zero. Hence, net force on the particle will be zero i.e., the particle stays at rest in its original position.
A satellite in a circular orbit of radius r has time period of 4 hrs. A satellite with orbital radius of 4r around the same planet will have a time period of:
Radius of orbit of the first satellite
R_{1} = R
Time period of the first satellite T = 4 hrs. and radius of orbit of second satellite = 4r The time period of satellite is given by
T_{2} = 8 × 4 = 32 hours.
Radius of orbit of the first satellite
R_{1} = R
Time period of the first satellite T = 4 hrs. and radius of orbit of second satellite = 4r The time period of satellite is given by
T_{2} = 8 × 4 = 32 hours.
A particle of mass M is at a distance a from the surface of a thin spherical shell of equal mass and having radius a.
At centre, field is zero due to shell but non zero due to particle
∴ E_{centre} = due to particle
Potential at centre is non zero due to both shell and particle.
At centre, field is zero due to shell but non zero due to particle
∴ E_{centre} = due to particle
Potential at centre is non zero due to both shell and particle.
A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F_{1} on a particle placed at a distance 3R from the centre of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in the figure. The sphere with cavity now applies a gravitational force F_{2} on the same particle. The ratio F_{2}/F_{1} is
From superposition principle, F_{1} = F_{r} + F_{c}
Here, F_{r} = force due to remaining part = F_{2}
And F_{c} = force due to cavity
From superposition principle, F_{1} = F_{r} + F_{c}
Here, F_{r} = force due to remaining part = F_{2}
And F_{c} = force due to cavity
If the distance between the earth And the sun were half its present value the number of days in a year would have been
T^{2} ∝ R^{3}
T^{2} ∝ R^{3}
A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg, mass of the earth = 6.0 × 10^{24} kg, radius of the earth = 6.4 × 10^{6} m, G = 6.67 × 10^{11} Nm^{2}/kg^{2}.
Given, height of the satellite above the earth’s surface (h) = 400 km = 0.4 × 10^{6} m
Mass of the satellite (m) = 200 kg
Radius of earth (R_{e}) = 6.4 × 10^{6} m
Mass of earth (M_{e}) = 6.0 × 10^{24} kg
Gravitational constant (G) = 6.67 × 10^{11} Nm^{2}/kg^{2}
Energy required send a satellite out of earth’s gravitational influence is called its binding energy.
Binding energy of a satellite =
= 5.885 × 10^{9} J
= 5.9 × 10^{9} J
Given, height of the satellite above the earth’s surface (h) = 400 km = 0.4 × 10^{6} m
Mass of the satellite (m) = 200 kg
Radius of earth (R_{e}) = 6.4 × 10^{6} m
Mass of earth (M_{e}) = 6.0 × 10^{24} kg
Gravitational constant (G) = 6.67 × 10^{11} Nm^{2}/kg^{2}
Energy required send a satellite out of earth’s gravitational influence is called its binding energy.
Binding energy of a satellite =
= 5.885 × 10^{9} J
= 5.9 × 10^{9} J
A particle of mass m is moved from A to B as shown in figure. Then potential energy of the particle
Outside the shell V_{0} =
Therefore, from A to surface v and hence the potential energy will decrease.
Inside the shell potential is constant. Hence, the potential energy is constant.
Outside the shell V_{0} =
Therefore, from A to surface v and hence the potential energy will decrease.
Inside the shell potential is constant. Hence, the potential energy is constant.
A satellite is in a circular orbit around the earth has kinetic energy E_{k}. Minimum amount of energy that is added so that it escapes the earth’s gravitational field is:
Total Mechanical energy (kinetic energy)
∴ TME = – E_{k}
For escape, TME = 0
i.e., If, E_{k} is provided then TME. Becomes Zero
Hence. The minimum amount of energy that is added so that it escapes the earth’s gravitational field is E_{k}.
Total Mechanical energy (kinetic energy)
∴ TME = – E_{k}
For escape, TME = 0
i.e., If, E_{k} is provided then TME. Becomes Zero
Hence. The minimum amount of energy that is added so that it escapes the earth’s gravitational field is E_{k}.
Assertion From a solid sphere of radius R, a hole of radius R/2 is cut as shown in figure. To find the magnitude of gravitational potential and gravitational field strength at O, we can directly subtract Potential due to hole (before removing) from potential due to whole sphere (before removing). The same can be done to find field strength at O, although potential is a scalar quantity and field strength is a vector quantity.
Reason In gravity, it is done like this. It makes no difference, whether the field strength is added/subtracted by vector method or by scalar method.
At a distance 320 km above the surface of earth, the value of acceleration due to gravity will be lower than its value on the surface of the earth by nearly (radius of earth = 6400 km)
Assertion A planet may orbit around a star either in orbit P or orbit Q. The speed of planet at O is same for both orbits.
Reason Planets orbit around a star with uniform velocity.
In circular path, star should lie at centre.
In circular path, star should lie at centre.
If the distance between the earth and the sun were half its present value, the number of days in a year would have been
From Kepler’s third law
T_{2} = 129 days
From Kepler’s third law
T_{2} = 129 days
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Students can check the previous years’ question papers and GATE MCQ to prepare for the upcoming examination. Through the question papers, the students will get to know the type of questions asked as well details about the weightage and division of topics in the syllabus.
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JEE Main Chemistry Chemical Kinetics Online Test. JEE Main Online Test for Chemistry Chemical Kinetics. JEE Main Full Online Quiz for Chemistry Chemical Kinetics. JEE Main Free Mock Test Paper 2019. JEE Main 2019 Free Online Practice Test, Take JEE Online Test for All Subjects. JEE Main Question and Answers for Chemistry Chemical Kinetics. In this test You may find JEE Main all subjects Questions with Answers. Check JEE Main Question and Answers in English. This mock Test is free for All Students. Mock Test Papers are very helpful for Exam Purpose, by using below Mock Test Paper you may Test your Study for Next Upcoming Exams. Now Scroll down below n Take JEE Main Chemistry Chemical Kinetics…
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A rapid reaction is distinguished by
In the following graphical representation for the reaction A → B there are two types of regions
The reaction of ozone with oxygen atom in the presence of chlorine atom can occur by the two – step process shown here with the rate constant for each step
O_{3}(g) + Cl(g) → O_{2}(g) + ClO(g), K_{1} = 5.2 × 10^{9} L mol^{–1} s^{–1}
ClO(g) + O(g) → O_{2}(g) + Cl(g), K_{2} = 2.6 × 10^{10} L mol^{–1} s^{–1}
Which of the values below is the closest to the rate constant of the overall net reaction, given by the equation
O_{3}(g) + O(g) → 2O_{2}(g)?
The inversion of cane sugar into glucose and fructose according to the equation
C_{12}H_{22}O_{11} + H_{2}O → C_{6}H_{12}O_{6} + C_{6}H_{12}O_{6}
Is an example of
A reaction follows the given concentration–time graph. The rate for this reaction at 20 s will be
The rate law for the reaction
RCl + NaOH → ROH + NaCl is given by Rate = k [RCl]. The rate of this reaction
RCI + NaOH → ROH + NaCI
Rate = k[RCl]
For this reaction, rate of reaction depends upon the concentration of RCl.
It means, the rate of reaction is halved by reducing the concentration of RCl by one half.
RCI + NaOH → ROH + NaCI
Rate = k[RCl]
For this reaction, rate of reaction depends upon the concentration of RCl.
It means, the rate of reaction is halved by reducing the concentration of RCl by one half.
The rate of the reaction, A + B_{2} → AB + B is directly proportional to the concentration of A and independent of concentration of B_{2}, hence, rate law is
For a gaseous reaction, the units of rate of reaction are
For zero^{th} order reaction
For the decomposition of N_{2}O_{5} at a particular temperature, according to the equations
2N_{2}O_{5} → 4NO_{2} + O_{2}
N_{2}O_{5} → 2NO_{2} + 1/2 O_{2}
The activation energies are E_{1} and E_{2} respectively then
Activation energy of a reaction is constant at constant temperature hence, E_{1} = E_{2}.
Activation energy of a reaction is constant at constant temperature hence, E_{1} = E_{2}.
Which represents first order reaction out of I, II and III?
Units of rate constant of first and zero order reactions in terms of molarity M unit are respectively
For nth order reaction
K = (mol L^{–1})^{1–n} s^{–1}
For first order reaction
Unit of k = s^{–1}
For zero order reaction
Unit of k = mol L^{–1} s^{–1} = M s^{–1}
For nth order reaction
K = (mol L^{–1})^{1–n} s^{–1}
For first order reaction
Unit of k = s^{–1}
For zero order reaction
Unit of k = mol L^{–1} s^{–1} = M s^{–1}
For the above reaction, if concentration of A is changed from 0.1 M to 0.2 M, rate becomes
For a second order reaction, 2A → Product, a straight line is obtained if we plot
The rate of a chemical reaction doubles for every 10°C rise in temperature. If the temperature is increased by 60°C, the rate of reaction increases by
If the temperature is increased by 60°C then 10° increase has been made 6 times and therefore, rate will increase by 2^{6} = 64 time.
If the temperature is increased by 60°C then 10° increase has been made 6 times and therefore, rate will increase by 2^{6} = 64 time.
If in the fermentation of sugar in an enzymatic solution that is initially 0.12 M the concentration of the sugar is reduced to 0.06 M in 10 h and to 0.03 M in 20 h, then order of the reaction is
0.12 M → 0.06 M 10h (50% reaction)
0.12 M → 0.03 M 20 h (75% reaction)
Time of 75% reaction = 2 × time of 50% reaction
Which is property of first order reaction
0.12 M → 0.06 M 10h (50% reaction)
0.12 M → 0.03 M 20 h (75% reaction)
Time of 75% reaction = 2 × time of 50% reaction
Which is property of first order reaction
If a certain reaction is first order with respect to A, second order with respect to B and zero order with respect to C then what is the order of reaction?
Rate law dx/dt = k[A]^{1}[B]^{2}[C]^{0} = k[A]^{1}[B]^{2}
Hence, order of reaction = 1 + 2 = 3
Rate law dx/dt = k[A]^{1}[B]^{2}[C]^{0} = k[A]^{1}[B]^{2}
Hence, order of reaction = 1 + 2 = 3
For a reaction A + B → C + D, if the concentration of A is doubled without altering the concentration of B, the rate gets doubled. If the concentration of B is increased by nine times without altering the concentration of A, the rate gets tripled. The order of the reaction is
For reaction, A + B → C + D
r_{1} = k [A]^{a}[B]^{b} … (i)
r_{2} = 2r_{1} = k[2A]^{a}[B]^{b} … (ii)
and r_{3} = 3r_{1} = k[A]^{a} [9B]^{b} … (iii)
from (i) and (ii) a = 1
form (i) and (iii), b = 1/2
∴ Order = a + b
= 1 + 1/2 × 3/2 = 1.5
For reaction, A + B → C + D
r_{1} = k [A]^{a}[B]^{b} … (i)
r_{2} = 2r_{1} = k[2A]^{a}[B]^{b} … (ii)
and r_{3} = 3r_{1} = k[A]^{a} [9B]^{b} … (iii)
from (i) and (ii) a = 1
form (i) and (iii), b = 1/2
∴ Order = a + b
= 1 + 1/2 × 3/2 = 1.5
For the reaction A → B, the rate expression is r = k[A]^{n}. When the concentration of A is doubled, the rate of reaction is quadrupled. The Value of n is
Let r = k[A]^{n} … (i)
When concentration is doubled then 4r = k(2A)^{n} … (ii)
Divide Eq. (ii) by (i)
4 = 2^{n}
∴ n = 2
Let r = k[A]^{n} … (i)
When concentration is doubled then 4r = k(2A)^{n} … (ii)
Divide Eq. (ii) by (i)
4 = 2^{n}
∴ n = 2
In the reaction, P + Q → R + S the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is
By increase in temperature by 10 K, the rate of reaction becomes double. How many times the rate of reaction will be if the temperature is increased from 303 K to 353 K?
By increasing 10 K temperature the rate of reaction becomes double. When temperature is increased from 303 K to 353 K. The rate increases in steps of 10° and has been made 5 times. Hence, the rate of reaction should increase 2^{5} times, i.e., 32 times.
By increasing 10 K temperature the rate of reaction becomes double. When temperature is increased from 303 K to 353 K. The rate increases in steps of 10° and has been made 5 times. Hence, the rate of reaction should increase 2^{5} times, i.e., 32 times.
Which of the following graphs represents exothermic reaction?
For exothermic reaction,
Energy of reactants > energy of product, which is the case given in (i).
For exothermic reaction,
Energy of reactants > energy of product, which is the case given in (i).
For a reaction between gaseous compounds,
2A + B → C + D
The reaction rate = k[A][B]. If the volume of the container is made 1/4 of the initial, then what will be the rate of reaction as compared to the initial rate?
When volume is reduced to one fourth, concentrations become four times. Hence, the rate of reaction becomes 16 times as compared to the initial rate.
When volume is reduced to one fourth, concentrations become four times. Hence, the rate of reaction becomes 16 times as compared to the initial rate.
The rate constant of a reaction is given by k = 2.1 × 10^{10} exp (– 2700 RT). It means that
k = 2.1 × 10^{10} exp (– 2700/RT)
i.e., log k vs 1/T will be straight line
intercept of log k axis = log 2.1 × 10^{10}
k = 2.1 × 10^{10} exp (– 2700/RT)
i.e., log k vs 1/T will be straight line
intercept of log k axis = log 2.1 × 10^{10}
Which of the reactions represented in these diagrams will show the greatest increase in rate for a given increase in temperature?
For a single step reaction,
A + 2B → Products, the molecularity is
Pieces of wood burn faster than a log of wood of the same mass because
Given the following diagram for the reaction
A + B → C + D
The enthalpy change and activation energy for the reverse reaction, C + D → A + B are respectively
follows first order reaction. (A) → Product Concentration of A, changes from 0.1 M to 0.025 M in 40 min. Find the rate of reaction of A when concentration of A is 0.01 M
dx/dt = k [A] = 0.03466 × 0.01
= 3.47 × 10^{–4} M min^{–1}
dx/dt = k [A] = 0.03466 × 0.01
= 3.47 × 10^{–4} M min^{–1}
For a given reaction.
2H_{2}O_{2} → 2H_{2}O + O_{2}
Graph between log [H_{2}O_{2}] and time t is of the type Thus, rate law is
Logarithm function is linear which is for first order reaction.
Log [H_{2}O_{2}]_{t} = – + log [H_{2}O_{2}]_{0} (constant)
Graph between log [H_{2}O_{2}]_{t} and t is a straight line.
Logarithm function is linear which is for first order reaction.
Log [H_{2}O_{2}]_{t} = – + log [H_{2}O_{2}]_{0} (constant)
Graph between log [H_{2}O_{2}]_{t} and t is a straight line.
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Strain energy is the
The neutral axis of the crosssection a beam is that axis at which the bending stress is
Euler’s formula holds good only for
The object of caulking in a riveted joint is to make the joint
A steel bar of 5 mm is heated from 15° C to 40° C and it is free to expand. The bar Will induce
A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum normal stress will be
Two shafts ‘A’ and ‘B’ transmit the same power. The speed of shaft ‘A’ is 250 r.p.m. and that of shaft ‘B’ is 300 r.p.m. The shaft ‘B’ has the greater diameter.
The stress induced in a body, when suddenly loaded, is __________ the stress induced when the same load is applied gradually.
If the slenderness ratio for a column is 100, then it is said to be a __________ column.
A masonry dam may fail due to
In order to prevent crushing of masonry at the base of the dam, the maximum stress should be __________ the permissible stress of the soil.
When a body is subjected to two equal and opposite pushes, as a result of which the body tends to reduce its length, the stress and strain induced is compressive.
The bending moment at a point on a beam is the algebraic __________ of all the moments on either side of the point.
The maximum diameter of the hole that can be punched from a plate of maximum shear stress 1/4th of its maximum crushing stress of punch, is equal to (where t = Thickness of the plate)
Two closely coiled helical springs ‘A’ and ‘B’ are equal in all respects but the number of turns of spring ‘A’ is half that of spring ‘B’ The ratio of deflections in spring ‘A’ to spring ‘B’ is
The deformation per unit length is called
Strain resetters are used to
When a rectangular beam is loaded transversely, the maximum compressive stress is developed on the
The point of contraflexure is a point where
The simply supported beam ‘A’ of length l carries a central point load W. Another beam ‘B’ is loaded with a uniformly distributed load such that the total load on the beam is W. The ratio of maximum deflections between beams A and B is
The maximum stress produced in a bar of tapering section is at
The energy stored in a body when strained within elastic limit is known as
In compression test, the fracture in cast iron specimen would occur along
The bending stress in a beam is __________ section modulus.
When shear force at a point is zero, then bending moment is __________ at that point.
When a bar is cooled to – 5°C, it will develop
The limit of eccentricity is based upon no tension condition.
In order to know whether a column is long or short, we must know its slenderness ratio.
Resilience is the
A concentrated load is one which
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According to principle of conservation of energy, the total momentum of a system of masses in any direction remains constant unless acted upon by an external force in that direction.
The friction experienced by a body, when in motion, is known as
Two balls of equal mass and of perfectly elastic material are lying on the floor. One of the ball with velocity v is made to struck the second ball. Both the balls after impact will move with a velocity
The term ‘force’ may be defined as an agent which produces or tends to produce, destroys or tends to destroy motion.
The coefficient of restitution for elastic bodies is one.
The velocity ratio in case of an inclined plane inclined at angle θ to the horizontal and weight being pulled up the inclined plane by vertical effort is
The range of projectile on a downward inclined plane is __________ the range on upward inclined plane for the same velocity of projection and angle of projection.
The angle of inclination of a vehicle when moving along a circular path __________ upon its mass.
A body of weight W is required to move up on rough inclined plane whose angle of inclination with the horizontal is α. The effort applied parallel to the plane is given by(where μ = tanφ = Coefficient of friction between the plane and the body.)
If the resultant of two equal forces has the same magnitude as either of the forces, then the angle between the two forces is
A smooth cylinder lying on its convex surface remains in __________ equilibrium.
Coefficient of friction is the ratio of the limiting friction to the normal reaction between the two bodies.
The total time taken by a projectile to reach maximum height and to return back to the ground, is known as time of flight.
The range of a projectile is maximum, when the angle of projection is
The mechanical advantage of a lifting machine is the ratio of
Static friction is always __________ dynamic friction.
A body will begin to move down an inclined plane if the angle of inclination of the plane is __________ the angle of friction.
When a particle moves along a circular path with uniform velocity, there will be no tangential acceleration.
The bodies which rebound after impact are called
The maximum frictional force, which comes into play, when a body just begins to slide over the surface of the other body, is known as
The distance, between the point of projection and the point where the projectile strikes the ground, is known as range.
The algebraic sum of the resolved parts of a number of forces in a given direction is equal to the resolved part of their resultant in the same direction. This is known as
The triangle law of forces states that if two forces acting simultaneously on a particle, be represented in magnitude and direction by the two sides of a triangle taken in order, then their resultant may be represented in magnitude and direction by the third side of a triangle, taken in opposite order.
The angle between two forces when the resultant is maximum and minimum respectively are
The path of the projectile is a parabola.
A redundant frame is also called __________ frame.
A resultant force is a single force which produces the same effect as produced by all the given forces acting on a body.
When a rigid body is suspended vertically, and it oscillates with a small amplitude under the action of the force of gravity, the body is known as
If two blocks of equal mass are attached to the two ends of a light string and one of the blocks is placed over a smooth horizontal plane while the other is hung freely after passing over a smooth pulley, then the two blocks will have some motion.
During elastic impact, the relative velocity of the two bodies after impact is __________ the relative velocity of the two bodies before impact.
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Convert the following binary number to decimal.
01011_{2}
Convert the binary number 1001.0010_{2} to decimal.
Decode the following ASCII message.
10100111010100101010110001001011001
01000001001000100000110100101000100
The voltages in digital electronics are continuously variable.
One hex digit is sometimes referred to as a(n):
Which of the following is the most widely used alphanumeric code for computer input and output?
If a typical PC uses a 20bit address code, how much memory can the CPU address?
Convert 59.72_{10} to BCD.
Convert 8B3F_{16} to binary.
Which is typically the longest: bit, byte, nibble, word?
Assign the proper odd parity bit to the code 111001.
Convert decimal 64 to binary.
Convert hexadecimal value C1 to binary.
Convert the following octal number to decimal.
17_{8}
How many binary digits are required to count to 100_{10}?
The BCD number for decimal 347 is ________.
The binary number for octal 45_{8} is ________.
The sum of 11101 + 10111 equals ________.
Convert the following binary number to decimal.
10011010_{2}