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SUPER 15 UP TET Level - 2 Test 1657
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SUPER 15 UP TET Level - 2 Test 1657
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  • Question 1/15
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    The temporary change in behaviour due to continuous exposure to stimuli is called____________.
    Solutions
    The correct answer is option A. i.e.., Habituation. Habituation is a form of non-associative learning and it decreases in response to a stimulus after repeated presentations. Habituation refers to a decrement in response as a result of continuous exposure to stimuli. Learning is the permanent change in behavior as a result of experience. Motivation is the process that guides and maintains goal-oriented behaviors.

  • Question 2/15
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    Which of the following is a component of academic achievement of children?
    Solutions
    The correct answer is option D, i.e., all of the above. Academic achievement of children should not be limited to obtaining high marks. Marks should be just a form of feedback which help them improve and motivate them. Important aspects of academic achievement of children includes: development of critical thinking, problem solving attitude, deep understanding pf conceptual knowledge, development of skills and learning attitude, etc. Academic achievement of students depends on the following factors: bio-social, affective and environmental.

  • Question 3/15
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    A student enjoys recognition in the class. He/She enjoys power, status, attention and admiration. Which type of need of the children is satisfied here?
    Solutions
    Esteem need is of two types, first is the internal esteem need, that a person has to satisfy within himself eg- respect, confidence, freedom and achievement. External esteem needs of an individual include recognition by others, power, admiration and attention.
  • Question 4/15
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    सूफी सन्तों की शैली है
    Solutions

    * सूफी संतों की शैली 'मसनवी' है। जिस रचना के प्रारंभ में ईश वंदना, पैगंबर की स्तुति एवं शाहेवक्त की प्रशंसा की जाती है वह मसनवी शैली कहलाती है।

    * मसनवी फारसी साहित्य में एक प्रकार का ग्रंथ होता है। कई लेखकों की मसनवियाँ मिलती हैं।

  • Question 5/15
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    स्वर’ किसका भेद है?
    Solutions

     ‘स्वर’ वर्ण भेद है। हिन्दी भाषा में प्रयुक्त सबसे छोटी इकाई वर्ण कहलाती है। जैसे-अ, , , , , , क्, ख् आदि। उच्चारण और प्रयोग के आधार पर हिन्दी वर्णमाला के दो भेद किए गए हैं:

    (क) स्वर

    (ख) व्यंजन

  • Question 6/15
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    निम्न शब्द -युग्म में गलत अर्थ वाले विकल्प का चयन कीजिए -
    Solutions

    कपीश- हनुमान, सुग्रीव कपिश - मटमैला

    जघन्य - गर्हित, शूद्र जघन - नितम्ब

    नगर - शहर नागर - चतुर व्यक्ति, शहरी

    स्वर्ग - तीसरा लोक सर्ग - अध्याय

  • Question 7/15
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    Change the following sentence into an interrogative sentence:

    There was nobody to help me.

    Solutions

    The given sentence is an assertive sentence. Here, if there is ‘Nobody’ used in the assertive sentence, to make it interrogative sentence, ‘Nobody is replaced by ‘Who’. All other given options are inappropriate. So, the correct answer will be: Who was there to help me?

  • Question 8/15
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    Which of the following sentence has a request?
    Solutions

    A sentence that expresses a command or a request is called an imperative sentence. Here, in given sentence word ‘please’ shows request. All other given options are inappropriate. So, the correct answer will be: Could you pass this notebook to Shruti, please?

  • Question 9/15
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    Complete the following sentence with most appropriate choice.

    ‘He was very much ashamed _______ his mischief.’

    Solutions

    * The preposition 'with' means accompanied by another person or another thing.

    * The preposition 'of' is used to indicate relating to, belonging to.

    * The preposition 'over' means straight above something, but not touching as it is at a higher level.

    * The preposition 'for' is used to show direction, time, place, location, spatial relationships, to introduce an object. and to specify the use of something.

    * Hence, the correct answer is option B.

  • Question 10/15
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    If highest common factor of x2 + px – q and 5x2 – 3px – 15q is (x – 3), then value of p and q will be
    Solutions

    If highest common factor of x+ px – q and 5x2 – 3px – 15q is (x – 3)

    When x = 3

    Then

    x2 + px – q =0

    32 + 3p – q = 0

    3p – q = –9 …(1)

    Also,

    5x2 – 3px – 15q = 0

    ⇒ 5(3)2 – 3×3(x) – 15q = 0

    ⇒ 45 – 9p – 15q = 0

    ⇒ 9p + 15q = 45

    ⇒ 3p + 5q = 15 …(2)

    Subtracting equation (1) from (2) we get

    6q = 24

    ⇒ q = 4

    Putting q = 4 in equation (1) we get

    3p – 4 = –9

    p = –5/3

  • Question 11/15
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    The difference between the simple interest and compound interest for 3 years at % per annum is Rs. 394.240. Find the sum?

    Solutions

    The difference between the simple interest and compound interest for 3 years is Difference = P* r2* (r + 300)]/1003

    Where P = Principle=sum

    T = Time

    R = Rate of interest per annum

    SI = Simple Interest

    394.240 = [Sum*64*308]/1000

    Sum = (394.240*100*100*100)/(64*308)

    Sum = Rs. 20000

    Therefore Option A is the right answer.

  • Question 12/15
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    The HCF and LCM of two numbers are 13 and 1989 respectively. If one of the numbers is 117, then the other number is
    Solutions
    HCF = 13, LCM = 1989
    First number = 117 and
    Other number = ?
    We have,
    HCF × LCM = First Number × Other number
    13 × 1989= 117 × Other number
    Other number =
  • Question 13/15
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    Which scientist discovered the radioactive element radium?
    Solutions

    Marie Curie is the scientist who discovered the radioactive element radium. She was awarded the 1911 Nobel Prize in Chemistry for her discoveries and studies of the elements radium and polonium. She is the only woman so far, who has been awarded the Nobel Prize twice.

  • Question 14/15
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    The gas used in artificial ripening of green fruits is
    Solutions

    The gas used in artificial ripening of green fruits is acetylene. It acts like ethylene and accelerates the ripening process. Acetylene is the chemical compound with the formula CH. It is a hydrocarbon and the simplest alkyne. It is unstable in its pure form and thus is usually handled as a solution.

  • Question 15/15
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    Sound waves consist of 20 compressions and 20 rarefactions in 10 m length. The wavelength of the waves is
    Solutions
    We know that, 20 compressions and 20 rarefactions make 20  wave 

    Distance between rarefaction and compression is 10 m = 1000 cm.
         v= f *  λ
    1000= 20 * λ
    λ= 1000/20
    λ= 50 cm
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