Physics
General
Easy

Question

A body of mass m crosses the top most point of a vertical circle with critical speed. What will be tension in string when it is horizontal-

  1. mg  
  2. 2 mg  
  3. 3 mg  
  4. 6 mg  

The correct answer is: 3 mg

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Related Questions to study

General
physics

A particle of mass m is rotating by means of a string in a vertical circle. The difference in the tension at the bottom and top would be-

A particle of mass m is rotating by means of a string in a vertical circle. The difference in the tension at the bottom and top would be-

physicsGeneral
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biology

This artery passes blood to the kidney is:

Uricotelism is found in those animals, which need water conservation. It is the elimination of uric acid and urates as the main nitrogenous wastes in the form of paste or in a semisolid state

This artery passes blood to the kidney is:

biologyGeneral
Uricotelism is found in those animals, which need water conservation. It is the elimination of uric acid and urates as the main nitrogenous wastes in the form of paste or in a semisolid state
General
Biology

Bacteria are included in which of the following kingdoms

Bacteria are included in which of the following kingdoms

BiologyGeneral
General
Biology

3 semicircular canals located above utricle lie –

3 semicircular canals located above utricle lie –

BiologyGeneral
General
Biology

Read the two reaction A and B given below and select the correct option accordingly
A) ADP + Pi rightwards arrow ATP
B) ATP rightwards arrow ADP + Pi

We need to find the characteristic of the equation given


 

Read the two reaction A and B given below and select the correct option accordingly
A) ADP + Pi rightwards arrow ATP
B) ATP rightwards arrow ADP + Pi

BiologyGeneral
We need to find the characteristic of the equation given


 
General
Physics-

A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants K subscript 1 end subscript comma K subscript 2 end subscript blank a n d blank K subscript 3 end subscript as shown. If a single dieletric material is to be used to have the same capacitance C is this capacitors, then its dielectric constant K is given by

Capacitance of two capacitors each of area fraction numerator A over denominator 2 end fraction blank comma plate separation d but dielectric constants K subscript 1 end subscript and K subscript 2 end subscriptrespectively joined in parallel
C subscript 1 end subscript equals fraction numerator K subscript 1 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d divided by 2 end fraction plus fraction numerator K subscript 2 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d divided by 2 end fraction equals fraction numerator open parentheses K subscript 1 end subscript plus K subscript 2 end subscript close parentheses epsilon subscript 0 end subscript A over denominator d end fraction
It is in series with a capacitor of plate area A comma plate separation d divided by 2 blankand dielectric constantK subscript 3 end subscript blank i e comma blank C subscript 2 end subscript equals fraction numerator K subscript 3 end subscript epsilon subscript 0 end subscript A over denominator d divided by 2 end fraction.
If resultant capacitance be taken as C equals fraction numerator K subscript blank epsilon subscript 0 end subscript A over denominator d end fraction,
Then fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator C subscript 1 end subscript end fraction plus fraction numerator 1 over denominator C subscript 2 end subscript end fraction
therefore fraction numerator d over denominator K epsilon subscript 0 end subscript A end fraction equals fraction numerator d over denominator open parentheses K subscript 1 end subscript plus K subscript 2 end subscript close parentheses epsilon subscript 0 end subscript A end fraction plus fraction numerator d divided by 2 over denominator table row cell K subscript 3 end subscript epsilon subscript 0 end subscript A end cell row cell end cell end table end fraction
rightwards double arrow fraction numerator 1 over denominator K end fraction equals fraction numerator 1 over denominator K subscript 1 end subscript plus blank K subscript 2 end subscript end fraction plus fraction numerator 1 over denominator 2 K subscript 3 end subscript end fraction

A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants K subscript 1 end subscript comma K subscript 2 end subscript blank a n d blank K subscript 3 end subscript as shown. If a single dieletric material is to be used to have the same capacitance C is this capacitors, then its dielectric constant K is given by

Physics-General
Capacitance of two capacitors each of area fraction numerator A over denominator 2 end fraction blank comma plate separation d but dielectric constants K subscript 1 end subscript and K subscript 2 end subscriptrespectively joined in parallel
C subscript 1 end subscript equals fraction numerator K subscript 1 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d divided by 2 end fraction plus fraction numerator K subscript 2 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d divided by 2 end fraction equals fraction numerator open parentheses K subscript 1 end subscript plus K subscript 2 end subscript close parentheses epsilon subscript 0 end subscript A over denominator d end fraction
It is in series with a capacitor of plate area A comma plate separation d divided by 2 blankand dielectric constantK subscript 3 end subscript blank i e comma blank C subscript 2 end subscript equals fraction numerator K subscript 3 end subscript epsilon subscript 0 end subscript A over denominator d divided by 2 end fraction.
If resultant capacitance be taken as C equals fraction numerator K subscript blank epsilon subscript 0 end subscript A over denominator d end fraction,
Then fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator C subscript 1 end subscript end fraction plus fraction numerator 1 over denominator C subscript 2 end subscript end fraction
therefore fraction numerator d over denominator K epsilon subscript 0 end subscript A end fraction equals fraction numerator d over denominator open parentheses K subscript 1 end subscript plus K subscript 2 end subscript close parentheses epsilon subscript 0 end subscript A end fraction plus fraction numerator d divided by 2 over denominator table row cell K subscript 3 end subscript epsilon subscript 0 end subscript A end cell row cell end cell end table end fraction
rightwards double arrow fraction numerator 1 over denominator K end fraction equals fraction numerator 1 over denominator K subscript 1 end subscript plus blank K subscript 2 end subscript end fraction plus fraction numerator 1 over denominator 2 K subscript 3 end subscript end fraction
General
Physics-

A body weight 8 g when placed in one pan and 18 g when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty. The true weight of the body is

Let the mass of first and second pan be m1 and m2, respectively and their respective distance from the pivot be L1 and L2.

Let the true weight of the body be M.

As the beam is horizontal when both the pans are empty, thus net torque about the pivot is zero.
  m1gL1m2gL2=0
We get  m1L1=m2L2      ....(1)
The body weighs 8g in first pan.
  (M+m1)gL1=(8+m2)gL2
Or   ML1=8L2    ....(2)        (Using (1))
The body weighs 18g in second pan.
  (18+m1)gL1=(M+m2)gL2
Or   18L1=ML2    ....(3)        (Using (1))
Dividing equation (2) and (3) we get  18M=M8
Or  M2=8×18=144
  M=12 g 

A body weight 8 g when placed in one pan and 18 g when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty. The true weight of the body is

Physics-General
Let the mass of first and second pan be m1 and m2, respectively and their respective distance from the pivot be L1 and L2.

Let the true weight of the body be M.

As the beam is horizontal when both the pans are empty, thus net torque about the pivot is zero.
  m1gL1m2gL2=0
We get  m1L1=m2L2      ....(1)
The body weighs 8g in first pan.
  (M+m1)gL1=(8+m2)gL2
Or   ML1=8L2    ....(2)        (Using (1))
The body weighs 18g in second pan.
  (18+m1)gL1=(M+m2)gL2
Or   18L1=ML2    ....(3)        (Using (1))
Dividing equation (2) and (3) we get  18M=M8
Or  M2=8×18=144
  M=12 g 
General
Physics-

If a body of mass m is carried by a lift moving with an upward acceleration a, then the forces acting on the body are (i) the reaction R on the floor of the lift upwards (ii) the weight m g of the body acting vertically downwards. The equation of motion will be given by


a=0 Non Inertial frame

Also, for equation of motion, we need to make it inertial.
For this, we have to acceletare in opposite direction.
 Equation of motion  is:R=mg+ma

If a body of mass m is carried by a lift moving with an upward acceleration a, then the forces acting on the body are (i) the reaction R on the floor of the lift upwards (ii) the weight m g of the body acting vertically downwards. The equation of motion will be given by

Physics-General

a=0 Non Inertial frame

Also, for equation of motion, we need to make it inertial.
For this, we have to acceletare in opposite direction.
 Equation of motion  is:R=mg+ma
General
Chemistry-

Assertion : First law of thermodynamics is not adequate in predicting the direction of a process
Reason : Total energy of universe is constant.

If both (A) and (R) are true but (R) is not the correct explanation of (A).

Assertion : First law of thermodynamics is not adequate in predicting the direction of a process
Reason : Total energy of universe is constant.

Chemistry-General
If both (A) and (R) are true but (R) is not the correct explanation of (A).
General
biology

Which one of the following pairs is mismatched?

Pearl is obtained from pearl oyster (Pinctadavulgaris), while honey from Apisindica, lac from Kenialacca and silk from Bombyxmori.

Which one of the following pairs is mismatched?

biologyGeneral
Pearl is obtained from pearl oyster (Pinctadavulgaris), while honey from Apisindica, lac from Kenialacca and silk from Bombyxmori.
General
botany

Statement 1: Libriformfibres are true fibres.
Statement 2: Libriformfibres develop from non-functional tracheids by reduction.

When the fibres possess very thick walls and reduced simple pits they are known as libriform wood fibres because of their similarity to phloem fibres. Libriform wood fibres chiefly occur in woody dicotyledons. There are many transitional forms between fibres and normal tracheids, these transitional forms are designated as fibretracheids. A line of demarcation cannot be drawn between tracheids and fibretracheids

Statement 1: Libriformfibres are true fibres.
Statement 2: Libriformfibres develop from non-functional tracheids by reduction.

botanyGeneral
When the fibres possess very thick walls and reduced simple pits they are known as libriform wood fibres because of their similarity to phloem fibres. Libriform wood fibres chiefly occur in woody dicotyledons. There are many transitional forms between fibres and normal tracheids, these transitional forms are designated as fibretracheids. A line of demarcation cannot be drawn between tracheids and fibretracheids
General
botany

Statement 1: In collateral vascular bundles, the phloem is situated towards inner side.
Statement 2:In monocot stem, cambium is present.

In collateral vascular bundles, phloem is situated towards outer side and xylem towards inner side and both are found on same radii, but in monocot stem, vascular bundles are closed, i.e., cambium is absent.

Statement 1: In collateral vascular bundles, the phloem is situated towards inner side.
Statement 2:In monocot stem, cambium is present.

botanyGeneral
In collateral vascular bundles, phloem is situated towards outer side and xylem towards inner side and both are found on same radii, but in monocot stem, vascular bundles are closed, i.e., cambium is absent.
General
botany

Statement 1: Many organs of aquatic plants float in water.
Statement 2: Large air gaps are present in the collenchyma tissues of lotus leaf.

Many organs of aquatic plants float in water. The mesophyll between upper and lower epidermis is differentiated into palisade parenchyma and spongy parenchyma. Intercellular space is present among the spongy parenchyma cells.

Statement 1: Many organs of aquatic plants float in water.
Statement 2: Large air gaps are present in the collenchyma tissues of lotus leaf.

botanyGeneral
Many organs of aquatic plants float in water. The mesophyll between upper and lower epidermis is differentiated into palisade parenchyma and spongy parenchyma. Intercellular space is present among the spongy parenchyma cells.
General
botany

Statement 1: Apical and intercalary meristems contribute to the growth in length while the lateral meristems bring an increase in girth in maize.
Statement 2: Apical and intercalary meristems always increase the height of plants.

Apical and intercalary meristems always increase the height of plant and lateral meristem is responsible for secondary growth (increase in girth) but secondary growth does not occur in monocots, e.g., maize.
Hence, Assertion is false but Reason is true

Statement 1: Apical and intercalary meristems contribute to the growth in length while the lateral meristems bring an increase in girth in maize.
Statement 2: Apical and intercalary meristems always increase the height of plants.

botanyGeneral
Apical and intercalary meristems always increase the height of plant and lateral meristem is responsible for secondary growth (increase in girth) but secondary growth does not occur in monocots, e.g., maize.
Hence, Assertion is false but Reason is true
General
botany

Statement 1: All the endodermal cells of the root do not contain Casparian thickenings on their radial and transverse walls.
Statement 2: Passage cells are found in the root endodermis.

The characteristic feature of endodermal cells of roots is the presence of Casparian thickenings on their radial and transverse walls but all cells of endodermis do not have Casparian thickenings instead they have simple permeable cell wall. Those cells are called passage cells, which remain in contact with the protoxylem cells.

Statement 1: All the endodermal cells of the root do not contain Casparian thickenings on their radial and transverse walls.
Statement 2: Passage cells are found in the root endodermis.

botanyGeneral
The characteristic feature of endodermal cells of roots is the presence of Casparian thickenings on their radial and transverse walls but all cells of endodermis do not have Casparian thickenings instead they have simple permeable cell wall. Those cells are called passage cells, which remain in contact with the protoxylem cells.