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Subject | Number of questions |
---|---|

Basics of Science and Engineering (DDCET) | 50 |

Aptitude Test (Mathematics and Soft Skill) DDCET | 50 |

Total | 100 |

- Double displacement
- Decomposition
- Combination
- Displacement

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) Burning of coal involves the main component, carbon, reacting with oxygen \((O_2)\) in the air to form carbon dioxide\( (CO_2)\).

\(\Rightarrow\) This is a combination reaction where two elements or compounds combine to form a new substance.

- Combination
- Double displacement
- Decomposition
- None of above

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) Double displacement reactions are reactions in which the reactants undergo an exchange of ions.

\(\Rightarrow\) The example of a double displacement reaction is given below:

\(\quad \quad Na_2SO_4 + NaCl_2 \rightarrow BaSO_4 + 2NaCl\)

\(\Rightarrow\) Here \(Cl_2\) and \(SO_4\) are undergo exchange.

\(\Rightarrow\) This equation can be expressed in a generalized form as follows: \(AC + BD \rightarrow AD+ BC\)

- \(Zn+ 2H_2SO_4 \rightarrow 2ZnSO_4 + H_2\)
- \(2Zn+ H_2SO_4 \rightarrow ZnSO_4 + 2H_2\)
- \(Zn+ H_2SO_4 \rightarrow ZnSO_4 + H_2\)
- None of above

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) In the balanced equation, the quantity of atoms for each element stays constant, both before and after a chemical reaction.

\(\Rightarrow\) Here in option C, zinc \( (Zn)\) has 1 atom, hydrogen \(H\) has 2 atoms, sulphur \(S\) has 1 atom and oxygen \(O\) has 4 atoms on both the side. Hence that equation is balanced.

- Calcium nitrate
- Potassium phosphate
- Calcium chloride
- ) Calcium phosphate

**Correct Ans: D**

**Solution:**

\(\Rightarrow\) Tooth enamel acts as a highly durable shield for your teeth, composed of a tough substance called calcium phosphate that is even stronger than bone.

\(\Rightarrow\)This protective layer covers the visible part of your teeth

- \(KNO_3\)
- \(NH_4CL\)
- \(CH_3COONa\)
- \(Na_2CO_3\)

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) \(KNO_3\) is a salt formed by a strong acid \((HNO_3)\) and a strong base \((KOH)\).

\(\Rightarrow\) When strong acids and strong bases react, they tend to neutralize each other, resulting in a solution with a pH close to 7.

- Base + Water
- Salt+ Water
- Acid + Salt
- None of these

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) Metal oxides have a basic nature, so when an acid interacts with a metal oxide, they cancel each other out.

\(\Rightarrow\) This results in the formation of salt and water during the reaction.

- \(Na_2ZnO_3\)
- \(Na_2ZnO_2\)
- \(NaZnO_2\)
- \(Na_2Zn_2O_2\)

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) Sodium zincate is a chemical compound with the formula \(Na_2ZnO_2\).

\(\Rightarrow\) It is formed when zinc reacts with sodium hydroxide \((NaOH)\) in an aqueous solution.

\(\Rightarrow\) It is often used in galvanizing and electroplating processes and can also be found in some types of batteries.

- 9
- 7
- 2
- 5

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) The atomic number is 9. So it has 2 electrons in the innermost shell and 7 electrons in the outer shell.

- Applying grease
- Applying paint
- Applying a coating of zinc
- All of the above.

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) Applying a layer of zinc on an iron frying pan can prevent it from rusting.

\(\Rightarrow\) This technique is called Electroplating.

\(\Rightarrow\) As zinc is more reactive than iron, oxygen in the air reacts with the surface of zinc first, and iron remains intact.

- Gold
- Lead
- Both (A) and (B)
- None of these

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) Malleability defines material's capacity to undergo permanent deformation under compression without breaking.

\(\Rightarrow\) Gold has an atomic structure that enables it to be hammered into extremely thin sheets, which are so delicate that they appear almost transparent.

\(\Rightarrow\) This is commonly referred to as gold leaf.

- Number
- Character
- Label
- Date/time

**Correct Ans: C**

**Solution:**

Note - This question is canceled by the authority.

\(\Rightarrow\) In Excel, "label" is a word that's used for text entries that usually act as headers or identifiers in a spreadsheet, but it's not seen as a separate data type.

\(\Rightarrow\) "Character" is also not visible directly in the MS Excel data type. It is included in "text" data type.

- 400
- 300
- 200
- 100

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) Powerpoint can zoom in up to 400% and zoom out to a minimum of 10%.

- ltalic
- Magic tool
- Font
- Bold

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) There is no such tool named Magic tool in MS-Word.

\(\Rightarrow\) While, font, bold, italic are the tools available to chance the font styles.

- ALU + CU
- ROM + ALU
- RAM + ROM
- None of the above

**Correct Ans: A**

**Solution:**

\(\text {The CPU (Central Processing Unit) typically consists of two main components:}\)

\(\Rightarrow\) ALU (Arithmetic Logic Unit): Responsible for performing arithmetic and logical operations on data.

\(\Rightarrow\) CU (Control Unit): Manages the execution of instructions and coordinates the activities of the other hardware components.

- UNIX
- MS-DOS
- LINUX
- Microsoft Office

**Correct Ans: D**

**Solution:**

\(\Rightarrow\) Microsoft Office is not an operating system (OS).

\(\Rightarrow\) It is a suite of productivity software applications developed by Microsoft for use in offices and businesses.

- Lead
- Arsenic
- Mercury
- Nitrate

**Correct Ans: D**

**Solution:**

\(\Rightarrow\) Blue baby syndrome is often linked to nitrate pollution.

\(\Rightarrow\) Nitrate can pollute sources of drinking water, and when babies drink water that has high levels of nitrate, it can cause a condition where the blood's ability to carry oxygen is decreased.

\(\Rightarrow\) This can result in the skin turning a bluish color.

- Stratosphere
- Troposphere
- Mesosphere
- Thermosphere

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) Clouds are primarily present in the troposphere, which is the lowest layer of the Earth's atmosphere.

- Increase temperature
- Affect environment
- Deplete ozone
- Affect human body

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) CFCs release chlorine atoms when they are broken down by ultraviolet (UV) radiation in the stratosphere.

\(\Rightarrow\) These chlorine atoms then react with ozone \((O_3)\) molecules, causing the depletion of the ozone layer.

\(\Rightarrow\) The ozone layer is crucial for protecting life on Earth from harmful UV radiation.

- 5- 10%
- 15 -20%
- 20 -30%
- 40 - 50%

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) The approximate efficiency range of modern solar photovoltaic (PV) panels in converting sunlight into electricity varies, but it typically falls between 15% to 20%.

\(\Rightarrow\) However, some high-efficiency panels can achieve efficiencies upwards of 20% or even higher.

- Lead
- Benzene
- Particulate matter (PM10)
- Carbon Monoxide (CO)

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) Particulate matter (PM10) is primarily responsible for causing respiratory problems and cardiovascular diseases in humans.

\(\Rightarrow\) PM10 refers to inhalable particles with diameters that are generally 10 micrometers and smaller.

- Joule
- Volt
- Watt
- Ampere

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) The SI unit of power is the watt (W) which is equal to one joule per second.

- 0.1 mm
- 0.001 mm
- 1 mm
- 0.01 mm

**Correct Ans: D**

**Solution:**

**Given data:**

\(\Rightarrow\) Pitch distance of screw =0.5 mm

\(\Rightarrow\) Number of total division on circular scale = 50

\(\Rightarrow\) \(LC=\frac{pitch\:distance\:of\:screw}{number \:of\:total\:division\:on\:circular\:scale}\)

\(\Rightarrow\) \(LC=\frac{0.5}{50}=0.01\:mm\)

- \(10^{-5}\)
- \(10^{-4}\)
- \(10^{5}\)
- \(10^{7}\)

**Correct Ans: B**

**Solution:**

**Relation between newton (N) and dyne:**

\(1\: N=10^5\:dyne\)

\(1\:dyne=10^{-5}\:N\)

\(10\:dyne=10 \times10^{-5}\:N=10^{-4}\:N\)

- Newton
- Watt
- Joule
- Arg

**Correct Ans: The correct answer does not align perfectly with the provided options**

**Solution:**

\(\Rightarrow\) In the CGS (Centimeter-Gram-Second) system, the unit of energy is the erg. The erg is a small unit of energy.

\(\Rightarrow\) In the MKS (Meter-Kilogram-Second) system and SI (International System of Units), the unit of energy is the joule.

\(\Rightarrow\) The relationship between the erg and the joule:

\(1\) erg\(=10^{-7}\) joules \(\implies\) 1 joule \(= 10^7\) erg

- 0.01254
- 0.1254
- 1.254
- 0.001254

**Correct Ans: The correct answer does not align perfectly with the provided options**

**Solution:**

\(\Rightarrow\) Standard value of refractive index of water = 1.333

Absolute differences:

|1.30 - 1.333| = 0.033

|1.32 - 1.333| = 0.013

|1.34 - 1.333| = 0.007

|1.35 - 1.333| = 0.017

|1.31 - 1.333| = 0.023

Mean Absolute Error = (0.033+0.013+0.007+0.017+0.023) / 5= 0.093 / 5=0.0186

- Vernier caliper
- Micrometer screw
- Spherometer
- Physical balance

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) Vernier caliper is used to measure "Inner diameter of the ring.

- \(F=mv\)
- \(P=mv\)
- \(F=ma\)
- \(P=ma\)

**Correct Ans: C**

**Solution:**

** Newton's second law of motion:**

\(\Rightarrow\) Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

\(\Rightarrow\) Formula: \(F=m \times a\)

- \(0.6\:m/s^2\)
- \(6\:m/s^2\)
- \(6\:m/s\)
- \(0.6\:m/s\)

**Correct Ans: B**

**Solution:**

**Given data:**

\(\Rightarrow\) \(F= 30 \:N\)

\(\Rightarrow\) \(m=5 \:kg\)

Force acting on the object \(F=m \times a\)

\(a=F/m\)

\(a=30/5=6\:\:m/s^2\)

- China wares are wrapped in paper while packing.
- Bogies of a train are provided with the buffers.
- Both (A) and (B)
- None of above

**Correct Ans: C**

**Solution:**

**Impulse of Force:**

\(\Rightarrow\) Impulse of force is the change in momentum of an object and it depends on the force applied and the time duration of the force.

\(\Rightarrow\) Formula: Impulse of Force = force \(\times\) time

\(\Rightarrow\) SI unit: newton-second (N s)

\(\Rightarrow\) CGS unit: dyne-second.

\(\Rightarrow\) It is a vector quantity.

**Examples:**

1. China wares wrapped in paper while packing.

2. Bogies of a train are provided with buffers.

Therefore, from the above explanation the correct option is (C).

- Force of gravity
- Centripetal force
- Friction force
- Centrifugal force

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) A body moving in vertical circular motion experiences these three forces like Force of gravity, Centripetal force and Centrifugal force.

\(\Rightarrow\) A body moving in vertical circular motion does not experience Friction force.

- Gravitational force
- Magnetic force
- Electrical force
- All of above

**Correct Ans: D**

**Solution:**

\(\Rightarrow\) Field forces are any forces that act on an object without direct physical contact.

\(\Rightarrow\) There are three main types:1. Gravitational force, 2. Magnetic force, 3. Electrical force

Therefore, from the above explanation the correct option is (D) All of above.

- Total energy
- Kinetic energy
- Potential energy
- None of these

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) The work done by the external force on a system equals the change in total energy.

- \(10\: k\Omega\)
- \(15\: k\Omega\)
- \(20\: k\Omega\)
- \(24\: k\Omega\)

**Correct Ans: B**

**Solution:**

**Given data:**

\(\Rightarrow\) \(V=6\: V\)

\(\Rightarrow\) \(I=0.4\:mA=0.4 \times 10^{-3}\:A\)

\(V=I \times R\)

\(6=0.4 \times 10^{-3} \times R \implies R=15000\: \Omega=15\:k \Omega\)

- \( N m^2 /C^2\)
- \(A/m\)
- \(N /m^2C\)
- \(C^2 / N m^2\)

**Correct Ans: D**

**Solution:**

According to Coulomb's law

The magnitude of the electrostatic force \((F)\) between two point charges

\(F=k \frac{Q_1 Q_2}{d^2}\)

\(F=\frac{1}{4\pi \epsilon_0

} \frac{Q_1 Q_2}{d^2}\)

\(\epsilon_0= \frac{Q_1 Q_2}{4 \pi \times F \times d^2}=\frac{C^2}{Nm^2}\)

Hence, option (D) is correct.

- Electric field intensity
- Area
- Angle between the surface and the electric field.
- All of these

**Correct Ans: D**

**Solution:**

\(\Rightarrow\) The Electric flux depends on the Electric field intensity, Area and Angle between the surface and the electric field.

Electric flux (\(\phi\)) is a measure of the electric field passing through an area.

Calculation: \(\phi= E \times A \cos \theta\)

Where,

\(E\) - Magnitude of the electric field intensity

\(A\) - Area of the surface

\(\theta\) - Angle between the surface and the electric field.

Therefore, from the above explanation the correct option is (D) All of these.

- current
- specific resistance
- conductance
- voltage

**Correct Ans: C**

**Solution:**

\(\mathbf{Resistance:}\)

\(\Rightarrow\) Resistance of the conductor is the opposition to the flow of current. It is denoted by R.

\(\Rightarrow\) Unit : \(\Omega\)

\(\mathbf{Conductance:}\)

\(\Rightarrow\) Conductance is the offered by the conductor to the flow of current. It is denoted by G.

\(\Rightarrow\) Unit : \(\mho\) or S (siemens)

Hence, Inverse of Resistance is known as conductance means Conductance is the reciprocal of resistance (\(G=\frac{1}{R}\))

- \(10\:\Omega\)
- \(20\:\Omega\)
- \(30\:\Omega\)
- \(6.6\:\Omega\)

**Correct Ans: D**

**Solution:**

**Given data:**

\(\Rightarrow\) \(R_1=10\:\Omega\)

\(\Rightarrow\) \(R_2=20\:\Omega\)

Effective value of resistance (\(R_{eq}\)) when two resistors are connected in parallel combination

\(\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\)

\(R_{eq}=\frac{R_1 \times R_2}{R_1 +R_2}\)

\(R_{eq}=\frac{10 \times 20}{10 +20}=6.66\:\Omega\)

- Q/C
- C/V
- Q/V
- C

**Correct Ans: B**

**Solution:**

** Capacitance: **

\(\Rightarrow\) Capability of a capacitor to store electrical charge is called capacitance.

\(\Rightarrow\) Unit of capacitance is Farad or coulomb/volts and is denoted by F.

\(\Rightarrow\) \(Q \propto V,\:Q =CV,\:C =\frac{Q}{V}=\frac{coulomb}{volts}\) .

Therefore, from the above explanation the correct option is (B) \(C/V\)

- Iron, Aluminium, Copper, Silver
- Aluminium, Iron, Copper, Silver
- Copper, Iron, Silver, Aluminium
- Iron, Copper, Aluminium, Silver

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) The correct order of these metals in increasing order of thermal conductivity is:

Iron, Aluminium, Copper, Silver.

**Note:**

\(\Rightarrow\) Iron: Has a thermal conductivity of 50–80 W/m·K

\(\Rightarrow\) Aluminum: Has a thermal conductivity of 205–235 W/m·K

\(\Rightarrow\) Copper: Has a thermal conductivity of 390–400 W/m·K

\(\Rightarrow\) Silver: Has a thermal conductivity of 430 W/m·K

- \(40^\circ C\)
- \(40^\circ F\)
- \(-40^\circ C\) and \(-40^\circ F\)
- None of above

**Correct Ans: C**

**Solution:**

** Relation between Fahrenheit scale and Celsius scale:**

\(T_F=\frac{9}{5} T_C+32\)

Where,

\(T_F=\) Temperature on Fahrenheit scale

\(T_C=\) Temperature on Celsius scale

Consider \(T_C=-40^\circ C\), this value put in above equation

\(T_F=\frac{9}{5} \times(-40)+32\)

\(T_F=-72+32\)

\(T_F=-40^\circ F\)

- \(J/K\)
- \(Cal/K\)
- \(Cal/^\circ C\)
- All of above

**Correct Ans: D**

**Solution:**

** Heat Capacity **

\(\Rightarrow\) It is the ratio of the amount of heat \((Q)\) given to a body to a change in its temperature \((\Delta T)\).

\(\Rightarrow\) \(H_C = \frac{Q}{\Delta T}\)

\(\Rightarrow\) SI Unit: \(J/K\)

\(\Rightarrow\) Other Unit: \(Cal/K \:or\:J/^\circ C\:or\:Cal/^\circ C\)

Therefore, from the above explanation the correct option is (D) All of above.

- Decreased
- Increased
- Remain constant
- Zero

**Correct Ans: The correct answer does not align perfectly with the provided options**

**Solution:**

\(\Rightarrow\) The relationship between thermal conductivity and temperature varies depending on the material.

**Gases:** The thermal conductivity of gases rises as the temperature increases.

**Liquids:** Liquids typically exhibit a decrease in thermal conductivity as temperature increases, although there are some exceptions such as water, which shows an increase initially and then decreases.

**Solids:** Solids typically exhibit a decrease in thermal conductivity as temperature increases, primarily because higher temperatures lead to heightened lattice vibrations that impede the transfer of heat.

From the above explanation correct options are A and B

- \(C = Q / m \Delta T\)
- \(Q = C / m \Delta T\)
- \(H_c = Q / \Delta L\)
- \(Q = H_c / \Delta L\)

**Correct Ans: A**

**Solution:**

The mathematical formula for specific heat:

\(C=\frac{Q}{m\Delta T}\)

Where,

\(C-\) Specific heat

\(Q-\) Heat energy transferred

\(m-\) Mass of the substance

\(\Delta T-\) Change in temperature

- \(38^\circ C\)
- \(40^\circ C\)
- \(-40^\circ C\)
- \(-38^\circ C\)

**Correct Ans: B**

**Solution:**

** Relation between Fahrenheit scale and Celsius scale:**

\(T_F=\frac{9}{5} T_C+32\)

Where,

\(T_F=\) Temperature on Fahrenheit scale

\(T_C=\) Temperature on Celsius scale

\(T_F=\frac{9}{5} T_C+32\)

\(104=\frac{9}{5} T_C+32\)

\(104 \times 5=(9 \times T_C)+(32 \times 5)\)

\(T_C=\frac{(104 \times 5) -(32 \times 5)}{9}=40^\circ C\)

- Sound wave
- Light wave
- Gamma Ray
- Micro wave

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) Sound wave is not a type of electromagnetic wave; it is a mechanical wave that requires a medium (such as air, water, or solid material) for propagation.

- Longitudinal
- Transverse
- Both (A) and (B)
- None of the above

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) Light waves are classified as non-mechanical type of waves (transverse waves.).

\(\Rightarrow\) They do not require any medium for propagation, they can propagate through vacuum also.

**Note:**

** Longitudinal wave**

\(\Rightarrow\) The particles of the medium move parallel to the direction of propagation of wave.

\(\Rightarrow\) Examples: Sound waves, etc.

** Transverse wave **

\(\Rightarrow\) The particles of the medium move perpendicular to the direction of propagation of wave.

\(\Rightarrow\) Examples: Light waves (electromagnetic waves), Water surface waves, etc.

Therefore, from the above explanation the correct option is (B) Transverse.

- \(2000 \:m/s\)
- \(400 \:m/s\)
- \(200 \:m/s\)
- \(4000 \:m/s\)

**Correct Ans: D**

**Solution:**

**Given data:**

\(\Rightarrow\) \(f=2\:kHz=2000\:Hz\)

\(\Rightarrow\) \(\lambda=2\:m\)

\(v = f· \lambda \)

\(v = 2000 \times 2=4000\:m/s \)

** Note: **

** Frequency: **

\(\Rightarrow\) It is the number of oscillations or cycles of a wave that occur in a unit of time.

\(\Rightarrow\) It is denoted by \(f\)

\(\Rightarrow\) Unit: hertz (Hz), cycle per second (c/s)

**Wavelength: **

\(\Rightarrow\) It is defined as the distance between two successive points that are in phase.

\(\Rightarrow\) It is denoted by \(\lambda\)

\(\Rightarrow\) Unit: meter (m) or Angstrom unit \((1\:A^\circ=10^{-10}\:m)\)

- Coherent
- Monochromatics
- Divergent
- High intensity

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) Divergent is not a characteristic of LASER (Light Amplification by Stimulated Emission of Radiation).

- Total internal reflection
- Polarization
- Refraction
- Differaction

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) Optical fibre works on the principle of total internal reflection.

- 9 second
- 0.966 second
- 90 second
- 9.6 second

**Correct Ans: B**

**Solution:**

**Given data:**

\(\Rightarrow\) \(V=6000 \:m^3\)

\(\Rightarrow\) \(A=1000\:O.W.U\)

\(RT=\frac{0.161 \times 6000}{1000}=0.966\:seconds\)

**Sabine's formula for reverberation of time:**

\(RT=\frac{0.161 \times V}{A}\)

Where,

\(V\) is the volume of the room in cubic meters

\(A\) is the total absorption in the room in \(O.W.U.\)

\(RT\) is the reverberation time in seconds

- \(\log_e \frac{1}{2}\)
- \(\log_e 10\)
- \(\log_e 5\)
- \(\log_e 2\)

**Correct Ans: D**

**Solution:**

\(\lim_{x \rightarrow 0} (\frac{10^x-5^x}{x}) =\frac{0}{0}\)

This is of the form \(\frac{0}{0}\), so by L’Hospital’s rule and differentiate the numerator and denominator:

\(=\) \( \lim\limits_{x \to 0} (\frac{10^xlog_e10-5^xlog_e5}{1}) \)

\(=\) \( log_e10-log_e5 \)

\(=\) \( log_e5+log_e2 -log_e5 \)

\(=\) \( log_e2 \)

- \(\frac{1}{4}\)
- \(-\frac{1}{8}\)
- \(\frac{1}{6}\)
- \(-\frac{1}{4}\)

**Correct Ans: B**

**Solution:**

\(\lim_{n \rightarrow \infty} \frac{3n^2 - 11n -13}{(4n-5)(7-6n)}\)

Divide numerator and denominator by \(n^2\)

\(= \lim_{n \rightarrow \infty} \frac{3 - \frac{11}{n} -\frac{13}{n^2}}{(4-\frac{5}{n})(\frac{7}{n}-6)}\)

\(= \frac{3 - 0 -0}{(4-0)(0-6)}\)

\(= -\frac{1}{8}\)

- \(3\)
- \(-3\)
- \(6\)
- \(\frac{1}{3}\)

**Correct Ans: A**

**Solution:**

\(f(x)=log_4(x)\) then

\(f(64)=log_4(64)\)

\(f(64)=log_44^3\)

\(f(64)=3log_44\)

\(f(64)=3\)

- \(\sin 2x\)
- \(-\sin 2x\)
- \(\cos 2x\)
- \(-\cos 2x\)

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) For \(f(x) = \cos^2 x\),

\(\implies f^{'}(x) = \frac{d(\cos^2 x)}{dx} = 2 \cos x \: \frac{d}{dx} \cos x\)

\(\therefore f^{'}(x) =2 \cos x (-\sin x) = - \sin (2x)\)

- \(\frac{1}{\sqrt{x^2-9}}\)
- \(\frac{x}{\sqrt{x^2-9}}\)
- \(\frac{9}{{x^2-9}}\)
- \(\frac{x}{{x^2-9}}\)

**Correct Ans: D**

**Solution:**

\(\Rightarrow\) For \(f(x) =\log (\sqrt{x^2-9}) = \frac{1}{2} \log ({x^2-9}) \),

\(\implies f^{'}(x) = \frac{1}{2}[\frac{1}{x^2-9} \frac{d}{dx}(x^2-9)] \)

\(\implies f^{'}(x) = \frac{1}{2}[\frac{1}{x^2-9} \times 2x] \)

\(\implies f^{'}(x) = \frac{x}{x^2-9}\)

- \(\sec \theta\)
- \(\csc \theta\)
- \(\sin \theta\)
- \(\cos \theta\)

**Correct Ans: A**

**Solution:**

\(\Rightarrow x=\csc \theta \quad \therefore \frac{dx}{d \theta} =- \cot \theta \:\csc \theta \)

\(\Rightarrow y=\cot \theta \quad \therefore \frac{dy}{d \theta} =- \csc^2 \theta \)

\(\rightarrow \frac{dy}{dx} = \frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}} = \frac{- \csc^2 \theta}{- \cot \theta \:\csc \theta}\)

\(\therefore \frac{dy}{dx} = \sec \theta\)

- \(f^{"} (a) >0\)
- \(f^{"} (a) <0\)
- \(f^{'} (a) >0\)
- \(f^{'} (a) <0\)

**Correct Ans: A**

**Solution:**

Note - This question is canceled by the authority.

\(\Rightarrow\) If the function \(f(x)\) get that \(f^{"}(x)>0\), then the function will have minimum value.

\(\Rightarrow\) If the function \(f(x)\) get that \(f^{"}(x)<0\), then the function will have maximum value.

- \(10 \sin (10x-17)\)
- \(-10 \sin (10x-17)\)
- \(\frac{1}{10} \sin (10x-17)\)
- \(-\frac{1}{10} \sin (10x-17)\)

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) Let's consider, \(u=10x-17 \quad \implies du = 10 dx\)

\(\therefore \int \cos (10x-17) dx = \frac{1}{10}\int \cos u \:du \)

\(=\frac{1}{10} \sin u + c = \frac{1}{10} \sin (10x-17) + c\)

- \(-1\)
- \(0\)
- \(\frac{1}{2}\)
- \(\frac{1}{4}\)

**Correct Ans: B**

**Solution:**

\(\Rightarrow f(x) = \sin^5 x \cos^8 x\)

\(f(-x) = \sin^5 (-x) \cos^8 (-x) = - \sin^5 x \: \cos^8 x = - f(x)\)

\(\therefore\) The given function f(x) is an \(\text{Odd}\) function.

\(\Rightarrow\) The integral of an odd function over a symmetric interval about the origin is zero.

\(\therefore \int_{-1}^1 f(x) dx = 0\)

- \(e^x \csc^2 x\)
- \(-e^x \csc^2 x\)
- \(e^x \cot x\)
- \(-e^x \cot x\)

**Correct Ans: D**

**Solution:**

\(\Rightarrow\) As rule of integration by parts \(\int e^x [f(x)+f^{'} (x)] dx = e^x f(x) + c \quad ...(i)\)

\(Given : f(x) = (\csc^2 x- \cot x) = -(\cot x + (-\csc^2 x))\)

\(\rightarrow Consider, f(x) = \cot x \quad \implies f^{'} (x) = - csc^x \)

\(\rightarrow\) From equation (i),

\(\therefore \int e^x (\csc^2 x- \cot x) dx = - \int e^x [(\cot x + (-\csc^2x))] dx\)

\(\implies \int e^x (\csc^2 x- \cot x) dx = - e^x \cot x + c\)

- \(\frac{1}{2} \log \: \lvert x^2-6x+40\rvert \)
- \(-\frac{1}{2} \log \: \lvert x^2-6x+40\rvert \)
- \(2 \log \: \lvert x^2-6x+40\rvert \)
- \(-2 \log \: \lvert x^2-6x+40\rvert \)

**Correct Ans: A**

**Solution:**

\(\Rightarrow Let's \: consider\),

\(u = x^2-6x+40 \quad \implies du = (2x - 6) dx = \frac{1}{2} (x-3) dx\)

\(\therefore \int \frac{x-3}{x^2-6x+40} dx = \frac{1}{2} \int \frac{1}{u} \: du\)

\(= \frac{1}{2} ln \: \lvert u \rvert + c = \frac{1}{2} ln \lvert (x^2-6x+40) \rvert + c\)

- \(9\)
- \(\frac{1}{9}\)
- \(\frac{1}{2}\)
- \(-\frac{1}{2}\)

**Correct Ans: C**

**Solution:**

\(Given: \log_{\frac{1}{81}} \frac{1}{9} = \log_{\frac{1}{9^2}} \frac{1}{9}\)

\(= \log_{9^{-2}} 9^{-1} = \frac{-1}{-2} \:log_9 \:9 = \frac{1}{2}\)

- 9
- 10
- 11
- 12

**Correct Ans: D**

**Solution:**

\(Given: 4^{log_2 3} + 2^{log_8 27} = 2{^2}^{log_2 3} + {8^{\frac{1}{3}}}^{log_8 27}\)

\(= 2^{log_2 3^2} + 8^{log_8 27^{\frac{1}{3}}} = 2^{log_2 \: 9} +8^{log_8 \: 3} \)

\(=9 + 3 = 12\)

- 8
- 9
- 10
- 11

**Correct Ans: B**

**Solution:**

\(Observations: \text{19,15,12,k,8,3}\)

\(Mean = \frac{19+15+12+k+8+3}{6}\)

\(\implies 11 = \frac{57 + k}{6}\)

\(\implies 66 = 57 + k\)

\(\therefore k = 9\)

- \( \begin{bmatrix} 3 & -2 \\ 7 & 0 \end{bmatrix}\)
- \( \begin{bmatrix} 3 & 0 \\ -2 & 7 \end{bmatrix}\)
- \( \begin{bmatrix} 3 & -2 \\ 0 & 7 \end{bmatrix}\)
- \( \begin{bmatrix} 3 & 0 \\ 7 & -2 \end{bmatrix}\)

**Correct Ans: B**

**Solution:**

If \(A= \begin{bmatrix} 1 & -1 \\ 1 & 2 \end{bmatrix}\) and \(B= \begin{bmatrix} 2 & 1 \\ -1 & 3 \end{bmatrix}\) then :

\(AB = \begin{bmatrix} 1 & -1 \\ 1 & 2 \end{bmatrix} . \begin{bmatrix} 2 & 1 \\ -1 & 3 \end{bmatrix}\)

\(AB = \begin{bmatrix} (1)(2) +(-1)(-1) & (1)(1) +(-1)(3) \\ (1)(2) + (2)(-1) & (1)(1) + (2)(3) \end{bmatrix} \)

\(AB = \begin{bmatrix} 2 +1 & 1 -3 \\ 2 -2 & 1 + 6 \end{bmatrix} \)

\(AB = \begin{bmatrix} 3 & -2 \\ 0 & 7 \end{bmatrix} \)

Now, \((AB)^T = \begin{bmatrix} 3 & -2 \\ 0 & 7 \end{bmatrix} ^T\)

\((AB)^T = \begin{bmatrix} 3 & 0 \\ -2 & 7 \end{bmatrix}\)

- -3
- 3
- -4
- 4

**Correct Ans: D**

**Solution:**

For \(A= \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 0 \end{bmatrix} \)

\( |A| = 1((1)(0)-(3)(1)) -2((2)(0)-(3)(1)) +1((2)(1)-(1)(1))\)

\( |A| = 1(0-3) -2(0-3) +1(2-1)\)

\( |A| = -3 +6 +1\)

\( |A| = 4\)

- \( \frac{1}{5} \begin{bmatrix} 1 & 1 \\ -3 & 2 \end{bmatrix}\)
- \( \frac{1}{5} \begin{bmatrix} 2 & 1 \\ -3 & 1 \end{bmatrix}\)
- \( \frac{1}{5} \begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix}\)
- \( \frac{1}{5} \begin{bmatrix} -2 & 3 \\ -1 & -1 \end{bmatrix}\)

**Correct Ans: A**

**Solution:**

If \(A= \begin{bmatrix} 2 & -1 \\ 3 & 1 \end{bmatrix}\)

Adjoint of a Matrix \(A \) (two by two) can be found by changing values of diagonal elements and changing signs of off-diagonal elements.

\(Adj(A) = \begin{bmatrix}1 & 1 \\ -3 & 2 \end{bmatrix} \)

Determinant \(|A| = \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix}= 5\)

Inverse of matrix \(A\) is given by:

\( (A)^{-1} = \frac{Adj(A)}{|A|} = \frac{ \begin{bmatrix} 1 & 1 \\ -3 & 2 \end{bmatrix}}{|A|}\)

\( (A)^{-1} = \frac{1}{5} \begin{bmatrix} 1 & 1 \\ -3 & 2 \end{bmatrix} \)

- \( \begin{bmatrix} 4 & -3 \\ -1 & 6 \end{bmatrix}\)
- \( \begin{bmatrix} 4 & -3 \\ -1 & -6 \end{bmatrix}\)
- \( \begin{bmatrix} -1 & 2 \\ 4 & -9 \end{bmatrix}\)
- \( \begin{bmatrix} -1 & 2 \\ 4 & 9 \end{bmatrix}\)

**Correct Ans: D**

**Solution:**

If \(A= \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}\) and \(B= \begin{bmatrix} 1 & 0 \\ 2 & 3 \end{bmatrix}\) then :

\( \implies 3B -2A = 3 \times \begin{bmatrix} 1 & 0 \\ 2 & 3 \end{bmatrix} - 2\times \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \)

\( \implies 3B -2A = \begin{bmatrix} 3 & 0 \\ 6 & 9 \end{bmatrix} - \begin{bmatrix} 4 & -2 \\ 2 & 0 \end{bmatrix} \)

\( \implies 3B -2A = \begin{bmatrix} -1 & 2 \\ 4 & 9 \end{bmatrix} \)

- \(\frac{-1}{\sqrt{2}}\)
- \(\frac{1}{\sqrt{2}}\)
- \(\frac{\sqrt{3}}{2}\)
- \(\frac{-1}{2}\)

**Correct Ans: A**

**Solution:**

\(\sin{(225^\circ)} \)

\(= \sin{(180^\circ+45^\circ)}\)

\(=-\sin{45}\)

\(=-\frac{1}{\sqrt{2}}\)

- \(\frac{-12}{13}\)
- \(\frac{12}{13}\)
- \(\frac{5}{13}\)
- \(\frac{-5}{13}\)

**Correct Ans: C**

**Solution:**

\(sec^2\theta-tan^2\theta=1\)

\(sec^2\theta=1+ tan^2\theta\)

\(sec^2\theta=1+ \frac{12}{5}^2\)

\(sec^2\theta=1+ \frac{144}{25}\)

\(sec^2\theta= \frac{169}{25}\)

\(sec\theta= \frac{13}{5}\)

\(\implies \cos\theta= \frac{5}{13}\)

- \(\frac{2\pi}{3}\)
- \(\frac{-\pi}{6}\)
- \(\frac{5\pi}{3}\)
- \(\frac{5\pi}{6}\)

**Correct Ans: D**

**Solution:**

Note - This question is canceled by the authority.

In \(\cos^{-1} (-\frac{\sqrt{3}}{2})\), where \((-\frac{\sqrt{3}}{2} <0)\)

\(\rightarrow\) It means \( \cos^{-1} (-\frac{\sqrt{3}}{2})\) lies in \(2^{nd}\) / \(3^{rd}\) quadrant.

Now \(\cos^{-1} (\frac{\sqrt{3}}{2}) =\frac{\pi}{6} \)

So, \(\cos^{-1} (-\frac{\sqrt{3}}{2})= (\pi + \frac{\pi}{6}) \: or \: (\pi - \frac{\pi}{6})\)

\(\cos^{-1} (-\frac{\sqrt{3}}{2})= ( \frac{7\pi}{6}) \: or \: (\frac{5\pi}{6})\)

- \(\frac{\pi}{3}\)
- \(\frac{\pi}{4}\)
- \(\frac{\pi}{6}\)
- \(\frac{\pi}{2}\)

**Correct Ans: B**

**Solution:**

\(|\overline{x} \times \overline{y}| = |\overline{x}| |\overline{y}| \sin{\theta_{xy}}\)

It is given that \(|\overline{x} \times \overline{y}| = |\overline{x}| = |\overline{y}|= \sqrt{2}\)

\(\implies \sqrt{2} =\sqrt{2} \sqrt{2} \sin{\theta_{xy}}\)

\(\implies \sin{\theta_{xy}} =\frac{1}{\sqrt{2}}\)

\(\implies \theta_{xy}=\frac{\pi}{4}\)

- \(\frac{-3}{4}\)
- \(\frac{3}{4}\)
- \(\frac{4}{3}\)
- \(\frac{-4}{3}\)

**Correct Ans: A**

**Solution:**

If vectors \(( 2,\: 3,\: a )\) and \(( 3,\: -1,\: 4 )\) are perpendicular then \( ( 2,\: 3,\: a ) \cdot ( 3,\: -1,\: 4 ) = 0 \)

\( \implies ( 2,\: 3,\: a ) \cdot ( 3,\: -1,\: 4 ) =0 \)

\( 6-3+4a = 0\)

\( 3+4a=0\)

\(a= \frac{-3}{4}\)

- 2
- -2
- 1
- -1

**Correct Ans: D**

**Solution:**

For the general equation of circle \( x^2+y^2+2gx+2fy +c =0 \), the center of the circle is:

\((-g,\: -f)\)

By comparing it with given equation we get:

\(g=0\), \(f=-3m\)

the center of the circle is given as \((0,\: -3)\):

\(\implies -f=3m=-3\)

\( \implies m=-1\)

- \(2x+y-1=0 \)
- \( x+2y-1=0 \)
- \( x+y=0 \)
- \( x+2y+1=0 \)

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) Two-point form of line joining two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:

\(\frac{y−y_1}{y2−y1} = \frac{x−x1}{x2−x1}\)

By placing given two points \((1, \: -1)\) and \((-1, \: 1)\), above equation can be written as:

\(\implies\) \( \frac{y−(-1)}{1−(-1)} = \frac{x−1}{(-1)−1}\)

\(\implies\) \( \frac{y+1}{2} = \frac{x−1}{-2}\)

\(\implies\) \( -2y-2 = 2x−2\)

\(\implies\) \( 2x+2y=0\)

\(\implies\) \(x+y=0\)

- Photo
- Salutation
- Date
- Signature

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) A photo is not typically a part of a standard business letter. Therefore, among the options provided, "Photo" is not a part of the business letter format.

\(\Rightarrow\) The other elements listed (Salutation, Date, Signature) are commonly included in a business letter.

- Signature
- Postscript
- Enclosure
- Photo

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) A letter is incomplete and worthless without a signature.

\(\Rightarrow\) The signature is a crucial element that indicates authenticity and authorship, confirming the sender's identity and agreement with the content of the letter. Therefore, "Signature" is the correct answer.

- Cash on delay
- Cash on delivery
- Cash on order
- Cash on distance

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) The full form of C.O.D. is "Cash on Delivery." It refers to a payment method where payment is made upon the delivery of goods rather than in advance.

- Examination
- Passed
- Hurrah!
- The whole sentence.

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) Interjections are words or phrases used to express emotions or sentiments and are often followed by an exclamation mark. In this case, "Hurrah!" expresses joy or excitement about passing the examination.

- under
- into
- between
- up

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) The appropriate option to complete the sentence is "under." Therefore, the sentence would be: "The dog is sitting under the cot." This option indicates the location of the dog in relation to the cot.

- raise
- rises
- rise
- raised

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) The appropriate option to complete the sentence is "rises." Therefore, the complete sentence is: "The Sun rises in the East." This option correctly conjugates the verb "rise" to agree with the subject "Sun" and the present tense of the sentence.

- are
- is
- has
- have

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) In this sentence, "Your family" is the subject, and it refers to a singular entity, even though it consists of multiple individuals. "Is" is the appropriate verb to use with a singular subject in the present tense.

- some
- any
- many
- much

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) The appropriate option to complete the sentence is "any." Therefore, the complete sentence is: "Do you have any doubt in this question paper?"

\(\Rightarrow\) The word "any" is used in questions to inquire about the existence or presence of something, especially when the answer could be affirmative or negative. In this context, it is asking whether there is doubt in the question paper, implying that the speaker is open to the possibility of either answer.

- Playwright
- Playwrite
- Playright
- Playrite

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) A "playwright" is a person who writes plays. The term "playwright" combines the word "play," referring to a dramatic work meant for performance on a stage, and "wright," which historically meant a craftsman or someone who creates something. So, a playwright is essentially someone who crafts or creates plays.

- Enterpeneur
- Entrepreneur
- Enterpreneur
- Entrepeneur

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) The correct spelling is "Entrepreneur."

- People were coming, going and ignorant him.
- People were coming, going and ignoring him.
- People where coming, going and ignoring him
- People were coming, going and ignored him.

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) The correct grammatical sentence is: "People were coming, going and ignoring him."

\(\Rightarrow\) This sentence uses the correct verb form "ignoring" to match the progressive tense of the sentence ("were coming, going"). The verb "ignoring" is in the present participle form, which is appropriate for indicating ongoing action in the past tense.

- I like drawing my coworkers and my cat.
- I Like drawing, my coworkers and my cat.
- I like drawing, my coworkers, and, my cat.
- I, Like, drawing, my coworkers, and, my, cat.

**Correct Ans: B**

**Solution:**

Note - This question is canceled by the authority.

\(\Rightarrow\) The grammatically correct sentence is:

I like drawing, my coworkers and my cat.

\(\Rightarrow\)

Here's why the other options are incorrect:

\(\rightarrow\) I Like drawing my coworkers and my cat. - No proper use of comma.

\(\rightarrow\) I like drawing, my coworkers, and, my cat. - This one has unnecessary commas.

\(\rightarrow\) I, Like, drawing, my coworkers, and, my, cat. - This one uses unnecessary commas and capitalization.

\(\Rightarrow\) In optionB, sentence consists of a simple subject ("I"), a verb ("like"), and a direct object ("drawing my coworkers and my cat"). This structure is clear, concise, and grammatically correct.

- The ship was wrecked, and every man, woman and child have drowned.
- The ship was wrecked, and every man, woman, and child are drowned.
- The ship was wrecked, and every man, woman and child is drowned.
- The ship was wrecked, and every man, woman and child was drowned.

**Correct Ans: D**

**Solution:**

\(\Rightarrow\) The correct grammatical sentence is: "The ship was wrecked, and every man, woman, and child was drowned."

\(\Rightarrow\) In this sentence: "The ship was wrecked" is in the past tense, which matches the action of the shipwreck.

"Every man, woman, and child" is a collective noun that refers to a singular group, so it requires a singular verb form, "was drowned," to match.

\(\Rightarrow\) Therefore, the correct form is "was drowned," making the sentence grammatically accurate.

The science of fireworks is technically called 'pyrotechnics' from the Greek word 'pyr' meaning fire and 'technics' meaning an art. Pyrotechnics includes not only fireworks but also whole range of devices that use similar materials and principles, from safety matches that we use every day to solid fuel rocket boosters of space shuttle. The household match is considered a special pyrotechnic device, as all the pyrotechnic effect heat, smoke, light, gas and sound are present in it.

Some historians say that 'black powder', the basic material used in fireworks, was invented in India, Shukranti, written more than two thousand years ago, has references to weapons similar to guns and projectile weapons. However, the Chinese are generally considered the pioneers of pyrotechnics. They are said to have developed, black powered more than one thousand years ago. It took at least two thousand years for the knowledge to spread to the West, and it was only in 1242 that an English Monk. Roger Bacon, revealed the formula for "black powder". He considered it such a dangerous substance that he wrote of it in a code language.

\(\mathbf{Ques.}\) As per some historians 'Black Powder was invented in _________.

- India
- China
- Greek
- None of the above

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) Black Powder was invented in India.

- a Chinese writer
- an Indian historian
- an English Monk
- a Greek philosopher

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) Roger Bacon was an English Monk.

- Shukranti
- Roger Bacon
- Chinese
- Greek

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) Roger Bacon wrote about 'Black Powder' in code language.

- One thousand years ago
- In 1242
- Two thousand years ago
- Modern times

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) Shukranti was written "Two thousand years ago."

- Indian language
- Greek language
- Chinese language
- None of the above

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) The word "pyrotechnics" is from the Greek language.

- written
- oral and written
- oral
- none of these

**Correct Ans: B**

**Solution:**

\(\Rightarrow\) Verbal communication refers to using words to convey information. This can be done through both speaking (oral) and writing.

- body language
- sign language
- paralanguage
- code language

**Correct Ans: C**

**Solution:**

\(\Rightarrow\) Tone of voice, voice quality, style of speaking, stress, and intonation are part of "paralanguage."

\(\Rightarrow\) Paralanguage refers to the non-verbal elements of speech, such as tone, pitch, volume, and rhythm, which convey meaning alongside spoken words.

- two way
- three way
- one way
- single way

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) The process of communication is a "two way" process. This implies that communication involves both sending and receiving messages between two or more parties.

- barrier
- noise
- sender
- receiver

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) The hindrance that disturbs the communication process at every stage is known as a "Barrier." Noise can be considered as one of the barrier.

- oral
- written
- verbal
- non-verbal

**Correct Ans: D**

**Solution:**

\(\Rightarrow\) Our physical appearance is an example of "non-verbal" communication. Non-verbal communication includes gestures, facial expressions, body language, posture, and other aspects of appearance that convey messages without the use of words.

- Sender's address
- Receiver's address
- Date
- Salutation

**Correct Ans: A**

**Solution:**

\(\Rightarrow\) At the top of a business letter on the left-hand side, you typically find the "Sender's address." This includes the sender's name, street address, city, state, and postal code.

- Just above the date
- Just below the date
- Along with the subject
- None of these

**Correct Ans: D**

**Solution:**

\(\Rightarrow\) The salutation in a business letter is typically placed "Just above the body of the letter." It is the greeting that addresses the recipient of the letter.