Free Snowflake ADA-C01 Exam Actual Questions

The questions for ADA-C01 were last updated On Feb 20, 2025

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Question No. 1

A Snowflake account is configured with SCIM provisioning for user accounts and has bi-directional synchronization for user identities. An Administrator with access to SECURITYADMIN uses the Snowflake UI to create a user by issuing the following commands:

use role USERADMIN;

create or replace role DEVELOPER_ROLE;

create user PTORRES PASSWORD = 'hello world!' MUST_CHANGE_PASSWORD = FALSE

default_role = DEVELOPER_ROLE;

The new user named PTORRES successfully logs in, but sees a default role of PUBLIC in the web UI. When attempted, the following command fails:

use DEVELOPER_ROLE;

Why does this command fail?

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Correct Answer: C

According to the Snowflake documentation1, creating a user with a default role does not automatically grant that role to the user. The user must be explicitly granted the role by the role owner or a higher-level role. Therefore, the USERADMIN role, which created the DEVELOPER_ROLE, needs to explicitly grant the DEVELOPER_ROLE to the new user PTORRES using the GRANT ROLE command. Otherwise, the user PTORRES will not be able to use the DEVELOPER_ROLE and will see the default role of PUBLIC in the web UI. Option A is incorrect because the DEVELOPER_ROLE does not need to be granted to SYSADMIN before user PTORRES can use the role. Option B is incorrect because the new role can take effect immediately after it is created and granted to the user, and does not depend on the USERADMIN role logging out. Option D is incorrect because the new role will not be affected by the identity provider synchronization, as it is created and managed in Snowflake.


Question No. 2

The following SQL command was executed:

Use role SECURITYADMIN;

Grant ownership

On future tables

In schema PROD. WORKING

To role PROD_WORKING_OWNER;

Grant role PROD_WORKING_OWNER to role SYSADMIN;

Use role ACCOUNTADMIN;

Create table PROD.WORKING.XYZ (value number) ;

Which role(s) can alter or drop table XYZ?

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Correct Answer: C

According to the GRANT OWNERSHIP documentation, the ownership privilege grants full control over the table and can only be held by one role at a time. However, the current owner can also grant the ownership privilege to another role, which transfers the ownership to the new role. In this case, the SECURITYADMIN role granted the ownership privilege on future tables in the PROD.WORKING schema to the PROD_WORKING_OWNER role. This means that any table created in that schema after the grant statement will be owned by the PROD_WORKING_OWNER role. Therefore, the PROD_WORKING_OWNER role can alter or drop table XYZ, which was created by the ACCOUNTADMIN role in the PROD.WORKING schema. Additionally, the ACCOUNTADMIN role can also alter or drop table XYZ, because it is the top-level role that has all privileges on all objects in the account. Furthermore, the SYSADMIN role can also alter or drop table XYZ, because it was granted the PROD_WORKING_OWNER role by the SECURITYADMIN role. The SYSADMIN role can activate the PROD_WORKING_OWNER role and inherit its privileges, including the ownership privilege on table XYZ. The SECURITYADMIN role cannot alter or drop table XYZ, because it does not have the ownership privilege on the table, nor does it have the PROD_WORKING_OWNER role.


Question No. 3

DatabaseA has a single schema called Schema1. This schema contains many tables and views. The ANALYST role has privileges to select from all objects in

Database

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Correct Answer: C, E

According to the Snowflake documentation, when a database or schema is cloned, the clone inherits all granted privileges on the clones of all child objects contained in the source object, such as tables and views. However, the clone of the container itself does not inherit the privileges granted on the source container. Therefore, the ANALYST role will have SELECT privilege on all tables and views in DatabaseA_clone.Schema1, but not USAGE privilege on the database or schema. The type of view (secure or non-secure) does not affect the cloning of privileges.


Question No. 4

An Administrator loads data into a staging table every day. Once loaded, users from several different departments perform transformations on the data and load it into

different production tables.

How should the staging table be created and used to MINIMIZE storage costs and MAXIMIZE performance?

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Correct Answer: B

According to the Snowflake documentation1, a transient table is a type of table that does not support Time Travel or Fail-safe, which means that it does not incur any storage costs for maintaining historical versions of the data or backups for disaster recovery. A transient table can be dropped at any time, and the data is not recoverable. A transient table can also have a retention time of 0 days, which means that the data is deleted immediately after the table is dropped or truncated. Therefore, creating the staging table as a transient table with a retention time of 0 days can minimize the storage costs and maximize the performance, as the data is only loaded and transformed once, and then deleted after the production tables are populated. Option A is incorrect because creating the staging table as an external table, which references data files stored in a cloud storage location, can incur additional costs and complexity for data transfer and synchronization, and may not provide the best performance for data loading and transformation. Option C is incorrect because creating the staging table as a temporary table, which is automatically dropped when the session ends or the user logs out, can cause data loss or inconsistency if the session is interrupted or terminated before the production tables are populated. Option D is incorrect because creating the staging table as a permanent table, which supports Time Travel and Fail-safe, can incur additional storage costs for maintaining historical versions of the data and backups for disaster recovery, and may not provide the best performance for data loading and transformation.


Question No. 5

A Snowflake Administrator needs to persist all virtual warehouse configurations for auditing and backups. Given a table already exists with the following schema:

Table Name : VWH_META

Column 1 : SNAPSHOT_TIME TIMESTAMP_NTZ

Column 2 : CONFIG VARIANT

Which commands should be executed to persist the warehouse data at the time of execution in JSON format in the table VWH META?

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Correct Answer: C

According to the Using Persisted Query Results documentation, the RESULT_SCAN function allows you to query the result set of a previous command as if it were a table. The LAST_QUERY_ID function returns the query ID of the most recent statement executed in the current session. Therefore, the combination of these two functions can be used to access the output of the SHOW WAREHOUSES command, which returns the configurations of all the virtual warehouses in the account. However, to persist the warehouse data in JSON format in the table VWH_META, the OBJECT_CONSTRUCT function is needed to convert the output of the SHOW WAREHOUSES command into a VARIANT column. The OBJECT_CONSTRUCT function takes a list of key-value pairs and returns a single JSON object. Therefore, the correct commands to execute are:

1. SHOW WAREHOUSES;

2. INSERT INTO VWH_META SELECT CURRENT_TIMESTAMP (), OBJECT_CONSTRUCT (*) FROM TABLE (RESULT_SCAN (LAST_QUERY_ID ()));

The other options are incorrect because:

*

A) This option does not use the OBJECT_CONSTRUCT function, so it will not persist the warehouse data in JSON format. Also, it is missing the * symbol in the SELECT clause, so it will not select any columns from the result set of the SHOW WAREHOUSES command.

* B) This option does not use the OBJECT_CONSTRUCT function, so it will not persist the warehouse data in JSON format. It will also try to insert multiple columns into a single VARIANT column, which will cause a type mismatch error.

* D) This option does not use the OBJECT_CONSTRUCT function, so it will not persist the warehouse data in JSON format. It will also try to use the RESULT_SCAN function on a subquery, which is not supported. The RESULT_SCAN function can only be used on a query ID or a table name.