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NEET Test 48
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NEET Test 48
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  • Question 1/15
    4 / -1

    A galvanometer of resistance 50 Q is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce alls deflection to 20 divisions, the resistance in series should be

    Solutions

    The resistance across the galvanometer is 50Ω. 
    Which is connected to a battery V=3V with a series resistance R = 2950Ω . On the basis of the information provided in the question the circuit will look like above figure. The current across the galvanometer will be: 

    Current for 30 divisions = 10-3

    Current for 20 divisions =

    In order to reduce all deflection to 20 divisions , the resistance R is added which is solved as: 

    R = 4450 Ω

  • Question 2/15
    4 / -1

    Two coils of self-inductances 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is :

    Solutions

    Self-inductance of coil 1 = 2 mH 
    Self-inductance of coil 2 = 8 mH 
    When the total flux associated with one coil links with the other i.e., a case of maximum flux linkage, then 

  • Question 3/15
    4 / -1

    A wire of resistance 12 nm-1 is bent to form a complete circle of radius 10 cm. The resistance between its two diametrically opposite points, A and B as shown in the figure, is

    Solutions

    The resistance between the two diametrically opposite points A and B can be shown as : 

    The circumference of the circle is given by : 

    Since the resistance distribution is divided into two section so, the resistance of each section will be represented as :

    The equivalent resistance will be given as : 

  • Question 4/15
    4 / -1

    A student measures the terminal potential difference (V) of a cell (of emf and internal resistance r) as a function of the current (I) flowing through it. The slope and intercept of the graph between V and I , then respectively, equal

    Solutions

    According to Ohm’s law: 

    V = IR where I is the current and R is the resistance present in the conductor 
    Suppose V is the terminal potential difference of a cell 
    And the cell has the emf = ε and the internal resistance = r then we can write: 
    V + Ir =  ε
    If I = 0 
    Then we can write, V = ε i.e. terminal potential difference is equal to the emf of the cell. 
    The slope will be given as:

    By, following the Ohm’s law.

  • Question 5/15
    4 / -1

    A galvanometer having a coil resistance of 60 Ωshows full scale deflection when a current of. 1.0 A passes through it. It can be converted into an ammeter to read currents up to 5.0 A by

    Solutions

    Given that the galvanometer has a coil of resistance, G= 60Ω 
    And it shows a full scale deflection when current I=1.0A is passed through it. 
    And we know that a galvanometer can be converted into an ammeter by connecting a resistor in parallel to the galvanometer coil. 
    Here , G = 60 Ω , Ig= 1A, I = 5A and we need to find the value of R 
    IgG = (I – Ig)R 
    Or R = IgG/(I – Ig
    By putting the respective values R becomes 
    R = 15Ω

  • Question 6/15
    4 / -1

    Wave theory of light states that

    Solutions

    Maxwell found wave theory of light and found out that light is transmitted in the form of electromagnetic waves. Electromagnetic waves transmit with same velocity equals to that of the light. Electromagnetic waves do not require any medium for propagation.

  • Question 7/15
    4 / -1

    The terms ‘a’ and ‘b’ in the below diagram are respectively 

    Solutions

    In this diagram, the point of maximum amplitude of wave is crest while of lowest amplitude is trough.

  • Question 8/15
    4 / -1

    Which of the following is an application of emulsion?

    Solutions

    Emulsion is widely used in food industry (mayonnaise, milk). They have medical and metallurgical applications. Digestion of fats in intestine occur with help of emulsion.

  • Question 9/15
    4 / -1

    Increase in temperature causes adsorption

    Solutions

    Adsorption process is exothermic hence adsorption decrease with increase in temperature.

  • Question 10/15
    4 / -1

    A monoatomic gas, X, adsorbed on a surface follows Langmuir adsorption isotherm. Plot of the fraction of surface coverage, θ against the concentration of the gas [X], for very low concentration of the gas, is described by the equation.

    Solutions

    The general formulae of Langmuir adsorption isotherm is 
    Θ=K[X]/1+K[X] 
    For very low concentration of the gas, 1>>>>K[X] and hence, 1+K[X]=1
    ∴θ=K[X]/1=K[X]

  • Question 11/15
    4 / -1

    The term bark is a non-technical term which excludes which of the following:

    Solutions

    All the tissue outside the vascular cambium constitutes of Bark.  Various kinds of cell layers that constitute the bark are Periderm that includes Phellogen, Phellem and Phelloderm, primary cortex, Pericycle, Primary and secondary Phloem. These all tissue are present outside the vascular cambium. The bark is formed early in the season is called early or soft bark and thus formed the ends of the season is called late or hard bark.

    • Thus, Bark excludes the term Secondary cortex
  • Question 12/15
    4 / -1

    Atactostele is found in the stem of

    Solutions

    Atactostele is the type where vascular bundles are scattered. Monocot stems show this feature. Hence the correct answer would be maize.

  • Question 13/15
    4 / -1

    Casparian strips are found on radial and inner walls of

    Solutions

    Root endodermis has casparian strips in the form of suberin depositions of the radial and inner walls of the cells. It is a feature of the root tissue system.

  • Question 14/15
    4 / -1

    Majority of angiosperm families possess:

    Solutions

    Most of the angiosperm families have anatropous ovules. In an anatropous condition, ovules are inverted during the development phase so that the micropyle lies very close to the hilum. 

  • Question 15/15
    4 / -1

    Which of the following conditions is not in accordance with autogamy?

    Solutions

    Autogamy is refers to self-pollination where pollination occurs within the same flower. Self-pollination takes place in cleistogamous flowers where anther and stigma lie very close to each other. In homogamy, the pollen is shed at the same time when the stigma becomes receptive resulting in intra floral self-pollination. In bud-pollination, stigma and pollens ripen at the same time inside the bud resulting in self-pollination.

    In dicliny condition, flowers are unisexual bearing either stamens or pistils. Fertilisation occurs in these flowers by cross pollination.

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