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A sum of money under compound interest doubles itself in 4 years. In how many years will it become 16 times itself? 

 

Option: 1

12 years


Option: 2

16 years


Option: 3

8 years


Option: 4

None of these


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Posted by

Kumar

A certain loan amounts, under compound interest, compounded annually earns an interest of Rs.1980 in the second year and Rs.2178 in the third year. How much interest did it earn in the first year?

Option: 1

Rs.1600


Option: 2

Rs.1800


Option: 3

 Rs.1900


Option: 4

None of these

 


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Posted by

Ram

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The distance between two stations A and B is 440 km. A train starts at 4 p.m. from A and move towards B at an average speed of 40 km/hr. Another train starts B at 5 p.m. and moves towards A at an average speed of 60 km/hr. How far from A will the two trains meet and at what time?

Option: 1

200,8 p.m.


Option: 2

300,9 p.m.


Option: 3

200,9 p.m.


Option: 4

300,8 p.m.


The sum of the distances covered by both trains should be equal to the total distance between the stations A and B: 40t+60(t−1)=44040t + 60(t - 1) = 44040t+60(t−1)=440

40t+60t−60=440
100t−60=440
100t=500
t=5

So, the trains will meet 5 hours after Train A starts from A.

Train A starts at 4 p.m. so 5 hours from 4 p.m. is 4 p.m. + 5 hours = 9 p.m.

The two trains will meet 200 km from A at 9 p.m.

Option 3: 200, 9 p.m.

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Posted by

Sara P

PQ is a tunnel. A dog sits at the distance of 5/ 11 of PQ from P. The train whistle coming from any end of the tunnel would make the dog run. If a train approaches P and dog runs towards P the train would hit the dog at P. If the dog runs towards Q instead, it would hit the dog at Q. Find ratio of speed of train and dog?

 

Option: 1

5:2


Option: 2

16:5


Option: 3

11:1


Option: 4

34:3


Option 3 - 11:1

########            P____5_____¶______6_____Q

"¶" is the dog located inside the tunnel PQ and #####... is a train approaching to it. 

Case 1 - Dog decides to reach point P 

Distance covered by Dog - 5 parts and train reaches at P

Case 2 - Dog decides to run towards point Q

At the time train reaches to point P, the dog covers 5 parts in direction of Q. Now the left out part covered by dog will be 1 part towards Q .

At the time dog covers 1 part and reaches Q, train covers whole 11 parts of tunnel and reaches from P to Q. 

So, 11 part of train is eual to the 1 part covered by the dog to reach Q. 

Ratio of their speeds - Train:Dog = 11:1  

 

 

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Posted by

Vaibhavi Gupta

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Read the passage below and answer the following questions.

Cheating is considered a criminal offence under the Indian Penal Code. It is done to gain profit or advantage from another person by using some deceitful means. The person who deceives another knows for the fact that it would place the other person in an unfair situation. Cheating as an offence can be made punishable under Section 420 of the IPC. Scope of Section 415 Cheating is defined under Section 415 of the Indian Penal Code as whoever fraudulently or dishonestly deceives a person to induce that person to deliver a property to any person or to consent to retain any property. If a person intentionally induces a person to do or omit to do any act which he would not have done if he was not deceived to do so and the act has caused harm to that person in body, mind, reputation, or property, then the person who fraudulently, dishonestly or intentionally induced the other person is said to cheat. Any dishonest concealment of facts that can deceive a person to do an act that he would not have done otherwise is also cheating within the meaning of this section. Essential Ingredients of Cheating requires · deception of any person. Fraudulently or dishonestly inducing that person to deliver any property to any person or to consent that any person shall retain any property; or · intentionally inducing a person to do or omit to do anything which he would not do or omit if he were not so deceived, and the act or omission causes or is likely to cause damage or harm to that person in body, mind, reputation or property.

Deceit– a tort arising from an untrue or false statement of facts which are made by a person, recklessly or knowingly, with an intention that it shall be acted upon by the other person, who would suffer damages as a result. 

Fraud – a false or untrue representation of the fact, that is made with the knowledge of its falsity or without the belief in its truth or a reckless statement that may or may not be true, with an intention to induce a person or individual to act independent of it with the result that the person acts on it and suffers damages and harm. In other words, it is a wrong act or criminal deception with an intention to result in financial or personal gain.

Question :- D went to a moneylender, Z, for the loan. D intentionally pledges the gold article with Z taking the loan. D knows that the article is not made of gold. After a few days, D leaves the village. Decide.

Option: 1

-


Option: 2

D has committed cheating as well as fraud 
 


Option: 3

D has not committed the offence of cheating 

 


Option: 4

D has committed an act which is an offence of cheating as well as the tort of deceit


As per section 415 of IPC, D has intentionally pledged the gold article with Z, his intention is clearly to deceive Z and cause him financial loss hence D would be liable for the offence of Cheating. Actions of D are also held liable as a tort of deceit.

Hence option (d) is correct. 

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Posted by

Rishabh Jain

As shown in the figure, in Young's double slit experiment, a thin plate of thicknesst=10\mu m and refractive index t=1.2 is inserted infront of slit S_{1}. The experiment is conducted in air(\mu =1) and uses a monochromatic light of wavelength \lambda =500\; nm. Due to the insertion of the plate, centralmaxima is shifted by a distance of x\beta _{0}.\beta _{0} is the fringe-width befor the insertion of the plate. The value of the x is ________.

Option: 1

4


Option: 2

__


Option: 3

__


Option: 4

__


Given\; t=10 \times 10^{-6} \mathrm{~m} \mu=1.2

\lambda=500 \times 10^{-9} \mathrm{~m}

When the glass slab inserted infront of one slit then the shift of central fringe is obtained by

\begin{aligned} & \mathrm{t}=\frac{\mathrm{n} \lambda}{(\mu-1)} \\ & \Rightarrow \quad 10 \times 10^{-6}=\frac{\mathrm{n} \times 500 \times 10^{-9}}{(1.2-1)} \\ & 10 \times 10^{-6}=\frac{\mathrm{n} \times 500 \times 10^{-9}}{0.2} \\ & \mathrm{n}=4 \end{aligned}

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Posted by

Info Expert 30

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A square shaped coil of area70\; cm^{2} having 600 turns rotates in a magnetic field of 0.4 \mathrm{wbm}^{-2}, about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field. If the coil completes 500 revolution in a minute, the instantaneous emf when the plane of the coil is inclined at 60^{\circ} with the field, will be ________ V.\left(\text { Take } \pi=\frac{22}{7}\right)

 

 

Option: 1

44


Option: 2

__


Option: 3

__


Option: 4

__


\text { Area (A) }=70 \mathrm{~cm}^2=70 \times 10^{-4} \mathrm{~m}^2

\begin{aligned} & \mathrm{B}=0.4 \mathrm{~T} \\ & \mathrm{f}=\frac{500 \text { revolution }}{60 \text { minute }}=\frac{500}{60} \frac{\mathrm{rev} .}{\mathrm{sec} .} \end{aligned}

Induced emf in rotating coil is given by

\begin{aligned} & \mathrm{e}=\mathrm{N} \omega \mathrm{BA} \sin \theta \\ & =600 \times 2 \times \frac{22}{7} \times \frac{500}{60} \times 0.4 \times 70 \times 10^{-4} \sin 30^{\circ} \\ & =600 \times 2 \times \frac{22}{7} \times \frac{500}{6} \times 0.4 \times 70 \times 10^{-4} \times \frac{1}{2} \\ & =44 \text { Volt } \end{aligned}

 

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Posted by

Gautam harsolia

A block is fastened to a horizontal spring. The block is pulled to a distance x = 10 cm from its equilibrium position (at x = 0 ) on a frictionless surface from rest. The energy of the block at x = 5 cm 0.25 J. The spring constant of the spring is ________Nm^{-1}

Option: 1

50


Option: 2

___


Option: 3

__


Option: 4

__


Given

A=10cm

At any instant total energy for free oscillation remains constant=\frac{1}{2} \mathrm{kA}^2

\begin{aligned} & \Rightarrow \frac{1}{2} \mathrm{kA}^2=0.25 \mathrm{~J} \\ & \Rightarrow \frac{1}{2} \mathrm{kA}^2=0.25 \mathrm{~J} \quad \Rightarrow \mathrm{K}=\frac{0.25 \times 2}{\mathrm{~A}^2} \\ & \Rightarrow \mathrm{k}=\frac{0.50}{(10 \mathrm{~cm})^2}=\frac{0.50}{\left(10 \times 10^{-2}\right)}=\frac{0.50 \times 10^4}{100} \\ & \mathrm{k}=0.50 \times 100=50 \mathrm{~N} / \mathrm{m} \end{aligned}

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Posted by

Divya Prakash Singh

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For a body projected at an angle with the horizontal from the ground, choose the correct statement.

Option: 1

The vertical component of momentum is maximum at the highest point.


Option: 2

The Kinetic Energy (K.E.) is zero at the highest point of projectile motion.


Option: 3

The horizontal component of velocity is zero at the highest point.


Option: 4

Gravitational potential energy is maximum at the highest point.


At highest point height is maximum and vertical component of velocity is zero.
So momentum is zero. At highest point horizontal component of velocity will not be zero but vertical component of velocity is equal to zero and because of this K.E. will not be equal to zero. Gravitational potential energy is maximum at highest point and equal to\mathrm{mgH}=\mathrm{mg}\left(\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\right)

Therefore the correct option is (4).

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Posted by

Gaurav

Nucleus A having \mathrm{Z=17} and equal number of protons and neutrons has \mathrm{1.2 \mathrm{MeV}} binding energy per nucleon. Another nucleus \mathrm{\mathrm{B}\: of\: Z=12} has total 26 nucleons and 1.8 \mathrm{MeV} binding energy per nucleons. The difference of binding energy of \mathrm{B \: and \: A} will be ___________\mathrm{\mathrm{MeV}.}
 

Option: 1

6


Option: 2

-


Option: 3

-


Option: 4

-


\mathrm{ \text { For Nucleus A } }
\mathrm{Z}=17=\text { Nummber of protons }           \mathrm{Given\left ( Z=N \right )\therefore N=17}
\mathrm{A}=34=\mathrm{Z}+\mathrm{N}
\mathrm{E}_{\mathrm{bn}}=1.2 \mathrm{MeV}
\frac{\mathrm{\left(\mathrm{E}_{\mathrm{B}}\right)_1 }}{A}=1.2 \mathrm{MeV}
\left(\mathrm{E}_{\mathrm{B}}\right)_1=(1.2 \mathrm{MeV}) \times \mathrm{A}
\left(\mathrm{E}_{\mathrm{B}}\right)_1=(1.2 \mathrm{MeV}) \times 34
\left(\mathrm{E}_{\mathrm{B}}\right)_1=40.8 \mathrm{MeV} \rightarrow \text { Binding energy of Nucleus } \mathrm{A}

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Posted by

Deependra Verma

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